It is tempting to write the equation as

[math]m(m^5-1)=n^2(n-5)[/math]

from which we derive a simple solution: [math](m,n)=(1,5)[/math]

From now on suppose [math]m\ge 2[/math]. By comparing magnitudes, we could guess that [math]n[/math] is approximately [math]m^2[/math]. So let’s write [math]n=m^2+d[/math]. Then [math]n^3-5n^2=m^6+(3d-5)m^4+(3d-10)dm^2+d^2(d-5)[/math], so

[math](3d-5)m^4+(3d-10)dm^2+d^2(d-5)=-m[/math]

It is immediate that there are no solutions if [math]d\ge 5[/math] and one can also verify easily the same holds if [math]d=4[/math] or [math]d=3[/math]. But if [math]d=2[/math] we have [math]m^4-8m^2-12=-m[/math]. Clearly, [math]m[/math] cannot be too large; in fact [math]m=4[/math] is already too large. But [math]m=3[/math] is a solution and since [math]m\ge 2[/math] the only one (in case [math]d=2[/math]). So we have another solution to the original equation:

[math](m,n)=(3,11)[/math]

The case [math]d=1[/math] is easily ruled out, for then [math]-2m^4-7m^2+m=4[/math] which forces [math]m[/math] to be a divisor of [math]4[/math], so [math]2[/math] or [math]4[/math], but those don’t work.

Finally, if [math]d\le 0[/math], then [math]n\le m^2[/math], so [math]n^2(n-5)\le m^4(m^2-5)=m^6-5m^2<m^6-m[/math]. So no solutions in this case and we have shown the only solutions to the original equation are [math](m,n)=(1,5),(3,11)[/math].

For a while let us forget the restriction [math] x \leq y \leq z. [/math]

Fix any prime number [math] p. [/math] For an integer [math] x, [/math] the additive valuation [math] v_p(x) [/math] at [math] p [/math] is the largest integer [math] c [/math] such that [math] p^{-c} x \in \mathbb{Z} [/math]. It satisfies

[math] v_p(a \pm b) \geq \min(v_p(a), v_p(b)) \tag{*} [/math]
where
[math](*) ~\text{holds with equality if }~ v_p(a) \neq v_p(b). \tag{**}[/math]

For [math] x, y, z \in \mathbb{Z}^+ [/math] satisfying the relation
[math] \mathrm{lcm}(x,y,z) = \gcd(x,y) + \gcd(y,z) + \gcd(z,x) \tag{0} [/math]
and a prime [math] p [/math] that divides at least one of [math] x, y, z, [/math] let us find cases that are impossible.

Suppose that [math] v_p(\gcd(x,y)) < \min(v_p(\gcd(y,z)), v_p(\gcd(z,x))). \tag{1} [/math]
Rewrite (0) as
[math]\mathrm{lcm}(x,y,z) - \gcd(x,y) = \gcd(y,z) + \gcd(z,x) \tag{2}[/math]

and evaluate the valuations at [math] p [/math] of both sides.
For LHS, by (*), (**) and the relation [math]v_p(\mathrm{lcm}(x,y,z)) \geq v_p(\gcd(y,z)) > v_p(\gcd(x,y)),[/math] it holds that
[math]v_p(\mathrm{lcm}(x,y,z) - \gcd(x,y)) = \min(v_p(\mathrm{lcm}(x,y,z)), v_p(\gcd(x,y)))[/math]

[math]\quad = v_p(\gcd(x,y))[/math]

on the other hand, for RHS, by (*), it holds that
[math]v_p(\gcd(y,z) + \gcd(z,x)) \geq \min(v_p(\gcd(y,z)),v_p(\gcd(z,x)) ), [/math]

thus, the assumption (1) contradicts the equality (0).

Therefore, among three nonnegative integers [math]v_p(\gcd(x,y)), v_p(\gcd(y,z)), v_p(\gcd(z,x)),[/math]

at least two of them are the smallest (i.e. the smallest can’t be alone).

Let us consider the case [math]v_p(\gcd(x,y)) = v_p(\gcd(y,z)) = v_p(\gcd(z,x)) [/math]first. This implies that [math] \min(v_p(x), v_p(y)) = \min(v_p(y), v_p(z)) = \min(v_p(z), v_p(x)), [/math]
which in turn implies that at least two of [math] v_p(x), v_p(y), v_p(z) [/math] are the smallest among these three.
By the condition [math] \gcd(x,y,z) = 1, [/math] the smallest value must be [math] 0. [/math] Also, since we have assumed that [math] p [/math] divides at least one of [math] x, y, z, [/math] The largest value should be positive.

The remaining case for [math]v_p(\gcd(x,y)), v_p(\gcd(y,z)), v_p(\gcd(z,x))[/math]is ``two equally small plus one big’’. WLOG, assume that
[math] v_p(\gcd(x,y)) = v_p(\gcd(y,z)) < v_p(\gcd(x,z)) \tag{3}. [/math] This implies that
[math]\min(v_p(x), v_p(y)) = \min(v_p(y), v_p(z)) < \min(v_p(z), v_p(x)) [/math] and thus [math] v_p(y) < \min(v_p(x), v_p(z)).[/math] By the condition [math] \gcd(x,y,z) = 1,[/math] we have [math] v_p(y) = 0. [/math]
Writing [math] x = p^a x’, z =p^b z’ [/math] with positive integers [math] a, b [/math] and [math] x’, z’ (v_p(x’) = v_p(z’) = 0),[/math] we have

[math] \mathrm{lcm}(x, y, z) = p^{\max(a,b)} \mathrm{lcm}(x’,y,z’) [/math]

[math]\geq p^{\max(a,b)} \mathrm{lcm}(x’,z’) \geq p^{\max(a,b)} \max(z’,x’)[/math]
and
[math] \gcd(x,y) + \gcd(y,z) + \gcd(z,x) [/math]

[math]\quad = \gcd(x’,y) + \gcd(y,z’) + p^{\min(a,b)} \gcd(z’,x’)[/math]

[math]\quad \leq x’ + z’ + p^{\min(a,b)} \gcd(z’,x’) [/math]

[math] \quad \leq 2 \max(z’,x’) + p^{\min(a,b)} \max(z’,x’). [/math]
Therefore, by (0), we have
[math] p^{\max(a,b)} \max(z’,x’) \leq 2 \max(z’,x’) + p^{\min(a,b)} \max(z’,x’) [/math]
which is simplified to
[math] p^{\max(a,b)} - p^{\min(a,b)} \leq 2. \tag{4}[/math]
Suppose further that [math] p > 2. [/math] Since [math] a [/math] and [math] b [/math] are positive,
[math] \max(a,b) = \min(a,b), [/math] or equivalently, [math] a = b [/math] should hold. In this case,
we have seen that [math] 0 = v_p(y) < a = v_p(x) = b = v_p(z). [/math]
When [math] p = 2, [/math] by dividing (4) by 2,
[math]2^{\max(a,b)-1} - 2^{\min(a,b)-1} \leq 1[/math]
holds. It is attained only when [math] a = b [/math] or [math]\max(a,b)=2 \wedge \min(a,b)=1.[/math]

Performing the above observations for each permutations of [math](x,y,z)[/math], we obtain the following lemma.

Lemma 1. Suppose that [math] x, y, z \in \mathbb{Z}^+ [/math] with [math] \gcd(x, y, z) = 1 [/math] satisfy the equation (0). For each prime [math] p > 2[/math] that divides at least one of [math] x, y, z, [/math] the set [math] \{ v_p(x), v_p(y), v_p(z) \}[/math] is [math] \{0, r\} [/math] where [math] r [/math] is a positive integer. For [math] p = 2,[/math] if at least one of [math] x, y, z [/math] is even, then [math] \{v_2(x), v_2(y), v_2(z) \} = \{0, 1, 2 \}[/math], or [math] = \{0, r\} [/math] where [math] r [/math] is a positive integer.

Let [math] S [/math] be the set of prime numbers [math] \geq 3 [/math] that divides at least one of [math] x, y, z. [/math] Lemma 1 asserts that [math] S [/math] is divided to the disjoint union
[math] S = S_{x} \sqcup S_{y} \sqcup S_{z} \sqcup S_{xy} \sqcup S_{yz} \sqcup S_{zx}, [/math]
where
[math] S_{x} := \{ p : v_p(x) > 0, v_p(y) = v_p(z) = 0 \}[/math]
[math] S_{xy} := \{ p : v_p(x) = v_p(y) > 0, v_p(z) = 0 \} [/math]
and so on. The prime factorizations of [math] x, y, z [/math] are:
[math] x = 2^{v_2(x)} \left(\prod_{p \in S_{x}} p^{v_p(x)} \right) \left(\prod_{p \in S_{xy}} p^{v_p(x)} \right) \left(\prod_{p \in S_{zx}} p^{v_p(x)} \right) [/math]

[math]\quad = 2^{v_2(x)} \left(\prod_{p \in S_{x}} p^{v_p(x)} \right) \left(\prod_{p \in S_{xy}} p^{v_p(y)} \right) \left(\prod_{p \in S_{zx}} p^{v_p(z)} \right), [/math]

[math]y = 2^{v_2(y)} \left(\prod_{p \in S_{y}} p^{v_p(y)} \right) \left(\prod_{p \in S_{xy}} p^{v_p(y)} \right) \left(\prod_{p \in S_{yz}} p^{v_p(y)} \right) [/math]

[math]\quad = 2^{v_2(y)} \left(\prod_{p \in S_{y}} p^{v_p(y)} \right) \left(\prod_{p \in S_{xy}} p^{v_p(x)} \right) \left(\prod_{p \in S_{yz}} p^{v_p(z)} \right), [/math]

[math]z = 2^{v_2(z)} \left(\prod_{p \in S_{z}} p^{v_p(z)} \right) \left(\prod_{p \in S_{yz}} p^{v_p(z)} \right) \left(\prod_{p \in S_{yz}} p^{v_p(z)} \right) [/math]

[math]\quad = 2^{v_2(z)} \left(\prod_{p \in S_{z}} p^{v_p(z)} \right) \left(\prod_{p \in S_{yz}} p^{v_p(y)} \right) \left(\prod_{p \in S_{zx}} p^{v_p(x)} \right). [/math]

If we write
[math] u_p := \begin{cases} v_p(x) & ~\text{if}~ p \in S_{x} \\ v_p(y) & ~\text{if}~ p \in S_{y} \\ v_p(z) & ~\text{if}~ p \in S_{z} \\ v_p(x) = v_p(y) & ~\text{if}~ p \in S_{xy} \\ v_p(y) = v_p(z) & ~\text{if}~ p \in S_{yz} \\ v_p(z) = v_p(x) & ~\text{if}~ p \in S_{zx}, \end{cases} [/math]

(note that it is well-defined) then a simplified presentation

[math] x = 2^{v_2(x)} \prod_{p \in S_{x} \sqcup S_{xy} \sqcup S_{zx} } p^{u_p} [/math]

[math]y = 2^{v_2(y)} \prod_{p \in S_{y} \sqcup S_{xy} \sqcup S_{yz} } p^{u_p} [/math]

[math] z = 2^{v_2(z)} \prod_{p \in S_{z} \sqcup S_{yz} \sqcup S_{zx} } p^{u_p} [/math]

and
[math] \gcd(x, y) = 2^{\min(v_2(x), v_2(y))} \prod_{p \in S_{xy}} p^{u_p} [/math]

[math] \gcd(y, z) = 2^{\min(v_2(y), v_2(z))} \prod_{p \in S_{yz}} p^{u_p} [/math]

[math]\gcd(z,x) = 2^{\min(v_2(z), v_2(x))} \prod_{p \in S_{zx}} p^{u_p} [/math]

[math] \mathrm{lcm}(x,y,z) = 2^{\max(v_2(x), v_2(y), v_2(z))} \prod_{p \in S} p^{u_p}[/math]

[math]\quad = 2^{\max(v_2(x), v_2(y), v_2(z))} w \prod_{p \in S_{xy} \sqcup S_{yz} \sqcup{zx} } p^{u_p},[/math]

where

[math]w := \prod_{p \in S_{x} \sqcup S_{y} \sqcup S_{z}} p^{u_p}[/math]

follow.

For the valuations at 2, first consider the case that one of [math] v_2(x), v_2(y), v_2(z)[/math] is 0 and the other two take a common value [math] t \geq 0 [/math]. *Edit* : [math] t > 0 [/math] is replaced with [math] t \geq 0 [/math] to cover the case [math] v_2(x) = v_2(y) = v_2(z) = 0. [/math]
Observe that [math] 2^{-t} \mathrm{lcm}(x,y,z) = 2^{-t} w \gcd(x,y) \gcd(y,z) \gcd(z,x). [/math]
Thus, (0) implies that
[math] w \gcd(x,y) \gcd(y,z) \gcd(z,x) = \gcd(x,y) + \gcd(y,z) + \gcd(z,x). [/math]
Put the multiset [math] \{e, f, g \} := \{\gcd(x,y), \gcd(y,z), \gcd(z,x) \} [/math] assuming
[math] e \geq f \geq g \geq 1. [/math] If [math] w f g > 3 [/math] then
[math] w e f g > 3 e \geq e + f + g. [/math] Thus [math] w f g \leq 3 [/math] is necessary for [math] w e f g = e + f + g. [/math]
When [math] w f g = 1, [/math] it is impossible that [math] e = e + f + g. [/math]
When [math] w f g = 2, [/math] since [math] w [/math] is odd by definition, it holds that[math] w = 1, \{f, g\} = \{1, 2\}, [/math] we have [math] 2 e = e + 1 + 2 [/math] which is attained by [math] e = 3. [/math] [math] \gcd(x,y) = 3, \gcd(y,z) = 2, \gcd(z,x) = 1, \mathrm{lcm}(x,y,z) = 6 [/math] is attained by [math] y = 6, x = 2, z = 3. [/math] Any permutation of [math] (6, 3, 2) [/math] is fine for the relation (0).
When [math] wfg = 3, [/math] [math] w [/math] is [math] 1 [/math] or [math] 3 [/math]. When [math] w = 1, [/math] by [math] \{f, g\} = \{1, 3\}, [/math] we have [math] 3 e = e + 1 + 3 [/math] and [math] e = 2. [/math] Thus, [math] \{e, f, g\} = \{1, 2, 3\} [/math] yields the same set of the solutions as the last case after the permutation. When [math] w = 3, [/math] we have [math] f = g = 1 [/math] and [math] 3 e = e + 1 + 1. [/math] Thus, [math] e = f = g = 1. [/math] [math]\gcd(x,y) = 1, \gcd(y,z) = 1, \gcd(z,x) = 1, \mathrm{lcm}(x,y,z) = 3 [/math] is attained by [math] x = 3, y = z = 1[/math](and any permutation of them is fine for (0)).

Next, consider the case that two of [math] v_2(x), v_2(y), v_2(z)[/math] are 0 and the other is [math] t > 0 [/math]. Observe that [math] 2^{-t} \mathrm{lcm}(x,y,z) = w \gcd(x,y) \gcd(y,z) \gcd(z,x). [/math]
Thus, (0) implies that
[math] 2^t w \gcd(x,y) \gcd(y,z) \gcd(z,x) = \gcd(x,y) + \gcd(y,z) + \gcd(z,x). [/math]
Again writing [math]\{e, f, g \} := \{\gcd(x,y), \gcd(y,z), \gcd(z,x) \} [/math] with
[math] e \geq f \geq g \geq 1, [/math]

[math]2^t w f g \leq 3 [/math]

is necessary. By [math] t > 0, [/math] the only possible case is [math] t = 1, w = f = g = 1. [/math]
In this case, [math] 2 e = e + 1 + 1, [/math] i.e., [math] e = 2 [/math] holds. [math]\gcd(x,y) = 2, \gcd(y,z) = 1, \gcd(z,x) = 1, \mathrm{lcm}(x,y,z) = 4[/math]
is attained by [math] (x, y, z) = (4, 2, 1). [/math] Any permutation is fine for (0). However, this violates the assumption that two of [math] v_2(x), v_2(y), v_2(z)[/math] are 0.

There remains the case [math] \{v_2(x), v_2(y), v_2(z) \} = \{0, 1, 2\}. [/math]
Observe that [math] 2^{-2} \mathrm{lcm}(x, y, z) = 2^{-0} 2^{-1} 2^{-0} w \gcd(x,y) \gcd(y,z) \gcd(z,x) [/math] or equivalently [math] \mathrm{lcm}(x, y, z) = 2 w \gcd(x,y) \gcd(y,z) \gcd(z,x) [/math] holds.
Again by writing [math] \{e, f, g \} := \{\gcd(x,y), \gcd(y,z), \gcd(z,x) \} [/math] with
[math] e \geq f \geq g \geq 1, [/math] (0) requires that [math] e + f + g = 2 w e f g [/math] which can be attained only when [math] 2 w f g \leq 3. [/math]In this case, it is forced that [math] w = f = g = 1. [/math]By the same calculation as the last case, [math](x, y, z) = (4, 2, 1) [/math] or its permutation. This time, [math]\{v_2(x), v_2(y), v_2(z) \} = \{0, 1, 2\}[/math] is attained.

Therefore, (0) is attained only by [math] (x, y, z) = (6,3,2), (4,2,1), (3, 1, 1) [/math] and their permutations. Recalling the constraint [math] x \leq y \leq z, [/math] the solution is [math] (x, y, z) \in \{ (2, 3, 6), (1, 2, 4), (1, 1, 3) \}. [/math]

[math]\boxed{[x,y,z]=[1,1,3],[1,2,4],[2,3,6]}[/math] are the only solutions.

Obviously [math]x=y=z[/math] is impossible because then [math]x=3x[/math]. If [math]x[/math], [math]y[/math] and [math]z[/math] are all co-prime to one another, [math]xyz=3[/math], which has a solution [math][x,y,z]=[1,1,3][/math] satisfying [math]x \le y \le z[/math]

Otherwise, [math]lcm(x,y,z) \le 2 \times max(gcd(x,y),gcd(y,z),gcd(z,x))+1[/math] because at least one of: [math]gcd(x,y)[/math], [math]gcd(y,z)[/math] and [math]gcd(z,x)[/math] is [math]1[/math].

(i) [math]x<y=z\implies xz \le 2z+1[/math]; this can only be satisfied when [math]x=[1,2][/math] because [math]y=z[/math] is at least [math]2[/math].

[math]x=1 \implies z=z+2[/math], which is impossible.

[math]x=2 \implies 2z=2+z[/math], which works only for [math]z=2[/math], which is impossible because [math]x<y=z[/math]

(ii) [math]x=y<z\implies xz \le 2x+1[/math]; because [math]z[/math] is at least [math]2[/math] and [math]x=y[/math] at least [math]1[/math], this can only be satisfied when [math]z=[2,3][/math]

[math]z=2 \implies x=y=1[/math]. This doesn't work because the equation in the question leads to [math]2 \ne 1+1+1[/math]

[math]z=3 \implies 3x \le 2x+1 \implies x=1 \implies y=1[/math]

[math][x,y,z]=[1,1,3][/math] works because [math]3=1+1+1[/math]

So, [math]x<y<z[/math] is all that is left.

(iii) [math]x<y<z[/math]:

a) [math]gcd(x,y)=1 \implies gcd(y,z) \le y[/math] and [math]gcd(z,x) \le x[/math] and [math]lcm(x,y,z) \ge xy[/math]

Comparing [math]xy[/math] and [math]1+gcd(y,z)+gcd(z,x) \le 1+x+y[/math] for [math]x<y[/math], only [math]x=[1,2][/math] would cause [math]xy \le 1+x+y[/math]. For [math]y>x \ge 3[/math], [math]xy>1+x+y[/math], so, [math]lcm(x,y,z)>1+gcd(y,z)+gcd(z,x)[/math]

[math]x=1 \implies lcm(x,y,z)=lcm(y,z)=2+gcd(y,z)[/math]. [math]lcm(y,z) \ge z[/math] and [math]gcd(y,z) \le y \implies 2+gcd(y,z) \le 2+y[/math]. To make something that is, at the maximum, [math]2+y[/math], equal something that is at least [math]z>y[/math], we need to only check the possibilities of [math]z=y+1[/math] and [math]z=y+2[/math].

[math]z=y+1[/math]:

[math]y[/math] and [math]z=y+1[/math] are co-prime, so [math]gcd(y,z)=1[/math], which means [math]lcm(y,z)=3 \implies [y,z]=[1,3][/math] for [math]y<z[/math]. But then [math]x<y \implies x<1[/math], which is impossible in this case. And [math]z \ne y+1[/math], as required.

[math]z=y+2[/math]:

if [math]y[/math] and [math]z[/math] are both odd, [math]gcd(y,z)=1[/math], which means [math]lcm(y,z)=3[/math], which means that [math][y,z]=[1,3][/math] for [math]y<z[/math]. Again, this is ruled out because [math]x<y<1[/math] is impossible.

if [math]y[/math] and [math]z[/math] are both even, [math]gcd(y,z)=2 \implies lcm(y,z)=1+2+1=4[/math], which, for [math]y[/math] and [math]z[/math] both even and [math]y<z \implies [y,z]=[2,4] \implies [x,y,z]=[1,2,4][/math], which works.

[math]x=2[/math]: [math]xy \le x+y+1 \implies 2y \le y+3[/math], which is only valid for [math]y=3[/math]; otherwise, for [math]y>4[/math], [math]2y>y+3[/math]. [math]z[/math] cannot be co-prime to both [math]x[/math] and [math]y[/math], so [math]z=2p[/math] or [math]3q[/math], where [math][p,q]>1[/math]

[math]lcm(x,y,z)\ge 2p[/math] or [math]3q[/math] and [math]1+gcd(y,z)+gcd(z,x) \le 6[/math] means that [math]2p[/math] or [math]3q=[4,6] \implies p=[2,3][/math]

[math]p=2 \implies z=4[/math]. But [math]lcm(2,3,4)=12>1+1+2[/math]

[math]p=3 \implies z=6[/math] and [math]lcm(2,3,6)=6=1+3+2=6[/math], works

[math]q=2 \implies z=6[/math] and the same [math][x,y,z]=[2,3,6][/math] works

b) [math]gcd(y,z)=1 \implies gcd(x,y) \le x[/math] and [math]gcd(z,x) \le x[/math]

[math]lcm(x,y,z)\ge yz[/math] and [math]1+gcd(x,y)+gcd(z,x) \le 1+2x[/math]

[math]yz \le 1+2x \implies z \le \frac{2x+1}{y}<2[/math] is impossible for [math]1\le x<y<z[/math], so no solutions exist.

c) [math]gcd(z,x)=1 \implies gcd(x,y) \le x[/math] and [math]gcd(y,z) \le y[/math]

[math]lcm (x,y,z) \ge xz[/math] and [math]1+gcd(x,y)+gcd(z,x) \le 1+x+y[/math]

[math]xz \le 1+x+y<1+2y \implies z \le \frac{1+2y}{x}[/math], which means that [math]x=1[/math]; for [math]x>1[/math], [math]z \le y[/math], which is impossible.

[math]x=1 \implies lcm(x,y,z)=lcm(y,z)=2+gcd(y,z)[/math]; this repeats the [math][x,y,z]=[1,2,4][/math] solution in a).

Assuming such a solution exists, multiplying both sides by [math]4[/math] and completing the square on the left side yields

[math]\begin{align*} (2x + 1)^2 &= 4y^4 + 4y^3 + 4y^2 + 4y + 1\\ &= (2y^2 + y + 1)^2 - (y^2 - 2y). \end{align*} \tag*{}[/math]

Then, we see that for all [math]y \geq 2[/math]:

[math](2y^2 + y)^2 \leq 4y^4 + 4y^3 + 4y^2 + 4y + 1 \leq (2y^2 + y + 1)^2, \tag*{}[/math]

where the left part of the inequality comes from

[math](4y^4 + 4y^3 + 4y^2 + 4y + 1) - (2y^2 + y)^2 = 3y^2 + 4y + 1 > 0. \tag*{}[/math]

Therefore when [math]y \geq 2[/math], we have bounded [math]4y^4 + 4y^3 + 4y^2 + 4y + 1[/math] between two consecutive perfect squares. This implies that [math]x[/math] can’t be an integer.

Thus, it only remains to check whether there is a positive integer solution for x when [math]y = 1, 2[/math]. By inspection, we find that the only solution in positive integers is

[math]\boxed{(x, y) = (5, 2)}. \tag*{}[/math]

[math]x^3+y^3=x^2+42xy+y^2\tag*{*)}[/math]

Glad you asked for HOW.

HOW.

1. Deal with x=y and in doing so finding the pair (22, 22).
2. Suppose that the GCD(x,y)=d and let x=dX and y=dY. Make the substitution and cancel out some d’s.
3. See if you can come up with [math]X^2-XY+Y^2[/math] divides [math]X^2+42XY+Y^2[/math].
4. Go one step farther to realize that [math]X^2-XY+Y^2[/math] must divide 43 which is nice since 43 is prime.
5. Deduce that [math]X^2-XY+Y^2 =1[/math] or [math]43[/math].
6. With [math]X^2-XY+Y^2–1=0[/math] viewed as a quadratic in X, insist that its discriminant be a square.
7. If [math]4–3Y^2[/math] is a square, there is only one acceptable value for Y and then find the corresponding X.
8. You should discover that in this case X and Y are equal which we dealt with in number 1.
9. With [math]X^2-XY+Y^2–43=0[/math] viewed as a quadratic in X, insist that its discriminant be square.
10. If [math]172–3Y^2[/math] is a square , there are only 7 values of Y to try. Try them
11. Having found all candidates for Y, deduce the corresponding X’s.
12. But we are not finished since x=dX and y=dY. So substitute into * and deal with equations only involving d and solve them.
13. Tidy up and never ask this question again.

How do I find all pairs [math](x,y)[/math] of positive real numbers such that [math]xy[/math] is an integer and [math]x+y=\lfloor x^2−y^2\rfloor[/math]?

Clearly [math]x+y[/math] is an integer, so let [math]w=x+y[/math]. Then [math]x^2-y^2=(x-y)(x+y)[/math]. Now let [math]d=x-y\ge0[/math] so that [math]2x=w+d[/math], [math]2y=w-d[/math].

From [math]x+y=\lfloor x^2-y^2\rfloor[/math], [math]0\le x+y\le(x-y)(x+y)<x+y+1[/math],

i.e. [math]0\le w\le dw<w+1[/math], i.e. [math]1\le d<1+\frac1w[/math].

For every integer [math]w[/math], there will be infinitely many values for [math]d[/math] in the above range.

But we also require that [math]xy[/math] be an integer, i.e. [math]4xy=w^2-d^2[/math] is a multiple of [math]4[/math] and, as [math]w[/math] is an integer, [math]d^2[/math] must be an integer.

If [math]w[/math] is even then [math]d[/math] must be an even integer, and if [math]w[/math] is odd then [math]w^2[/math] is odd and [math]d^2[/math] must be an odd integer. However, [math]d[/math] need not be an integer, and if [math]w>1[/math], it cannot be.

The case [math]w=1[/math] gives [math]d=1[/math] so [math]x=1[/math], [math]y=0[/math], which contradicts the given condition that [math]x>0[/math] and [math]y>0[/math].

If [math]w=2[/math] then [math]1\le d<1\frac12[/math], i.e. [math]1\le d^2<2\frac14[/math], so [math]d=1[/math] or [math]d^2=2[/math]. If [math]d=1[/math] then [math]x=1.5[/math], [math]y=1[/math] and [math]xy[/math] is not an integer; if [math]d^2=2[/math] then [math]2x=2+\sqrt2[/math], [math]2y=2-\sqrt2[/math]. But then [math]4xy=2[/math], so [math]xy[/math] is not an integer.

If [math]w\ge3[/math] then [math]1\le d<1\frac13[/math], i.e. [math]1\le d^2<1\frac79[/math], so [math]d=1[/math] and [math]x=\frac12(w+1)[/math], [math]y=\frac12(w-1)[/math]. In that case, [math]xy[/math] an integer only if [math]w[/math] is odd. In that case [math]x[/math] is an integer and [math]y=x-1[/math]. That is the obvious trivial solution.

So the solutions are [math]y=x-1[/math] where [math]x[/math] is an integer.

We want to solve the Diophantine equation:

[math]2^x+17 = y^4 … (1)[/math]

Evidently, [math]y[/math] must be odd. By working modulo[math]17[/math] for both sides, we have:

For the LHS:

[math]2^1 ≡ 2(mod17) … (2)[/math]

[math]2^2 ≡ 4(mod17) … (3)[/math]

[math]2^3 ≡ 8(mod17) … (4)[/math]

[math]2^4 ≡ 16(mod17) … (5)[/math]

[math]2^5 ≡ 15(mod17) … (6)[/math]

[math]2^6 ≡ 13(mod17) … (7)[/math]

[math]2^7 ≡ 9(mod17) … (8)[/math]

[math]2^8 ≡ 1(mod17) … (9)[/math]

[math]2^9 ≡ 2(mod17) … (10)[/math]

While for the RHS

For [math]m = 1 => (2m+1)^4 ≡ 13(mod17) … (11)[/math]

For [math]m = 2 => (2m+1)^4 ≡ 13(mod17) … (12)[/math]

For [math]m = 3 => (2m+1)^4 ≡ 4(mod17) … (13)[/math]

For [math]m = 4 => (2m+1)^4 ≡ 16(mod17) … (14)[/math]

For [math]m = 5 => (2m+1)^4 ≡ 4(mod17) … (15)[/math]

For [math]m = 6 => (2m+1)^4 ≡ 1(mod17) … (16)[/math]

For [math]m = 7 => (2m+1)^4 ≡ 16(mod17) … (17)[/math]

For [math]m = 8 => (2m+1)^4 ≡ 0(mod17) … (18)[/math]

For [math]m = 9 => (2m+1)^4 ≡ 16(mod17) … (19)[/math]

For [math]m = 10 => (2m+1)^4 ≡ 1(mod17) … (20)[/math]

For [math]m = 11 => (2m+1)^4 ≡ 4(mod17) … (21)[/math]

For [math]m = 12 => (2m+1)^4 ≡ 16(mod17) … (22)[/math]

For [math]m = 13 => (2m+1)^4 ≡ 4(mod17) … (23)[/math]

For [math]m = 14 => (2m+1)^4 ≡ 13(mod17) … (24)[/math]

For [math]m = 15 => (2m+1)^4 ≡ 13(mod17) … (25)[/math]

For [math]m = 16 => (2m+1)^4 ≡ 1(mod17) … (26)[/math]

For [math]m = 17 => (2m+1)^4 ≡ 1(mod17) … (27)[/math]

Hence, the only agreement for both sides is when:

[math]2^n[/math] is congruent to [math]1[/math], [math]4[/math], [math]13[/math], or [math]16(mod17)[/math] and this happens only if [math]x[/math] is even. Hence, there exists positive integer [math]p[/math], such that [math]x = 2p[/math], which leads to the equation:

[math]2^{2p}+17 = (y^2)^2 => (y^2)^2–(2^p)^2 = 17 => [/math]

[math](y^2–2^p)(y^2+2^p) = 17 … (28)[/math]

Now, since [math]17[/math] is prime and since [math]y^2+2^p[/math] is striclty positive and [math]y^2+2^p > y^2–2^p[/math], it follows that we can only have:

[math]y^2–2^p = 1 ... (29)[/math]

and:

[math]y^2+2^p = 17 ... (30)[/math]

By solving the system of equations [math](29)[/math] and [math](30)[/math], we easily obtain:

[math]2y^2 = 18 => y = 3[/math], therefore [math]2^p = 8 => p = 3[/math] and [math]x = 6[/math].

Hence, there is only one positive integral solution and this is:

[math](x[/math], [math]y) = (6[/math], [math]3)[/math]

Since

[math]x+y=\lfloor x^2-y^2 \rfloor[/math]

we also know that [math]x+y[/math] is integer. Lets set [math]x+y=n, xy=m[/math], with [math]n,m \in \mathbb{N}[/math]. From this we obtain

[math]x+\dfrac{m}{x}=n[/math]

or

[math]x^2-nx+m=0[/math]

or

[math](2x-n)^2+4m=n^2[/math]

Thus

[math]x=\dfrac{1}{2}\left(n \pm \sqrt{n^2–4m}\right)[/math]

and (matching signs!)

[math]y=\dfrac{1}{2}\left(n \mp \sqrt{n^2–4m}\right)[/math]

Making

[math]x-y=\pm \sqrt{n^2–4m}[/math]

And we find

[math]x^2-y^2=\pm n \sqrt{n^2–4m}[/math]

We thus need that

[math]\lfloor \pm n\sqrt{n^2–4m}\rfloor =n[/math]

Since [math]x,y[/math] are positive real numbers and [math]x+y=n \in \mathbb{N}[/math], we take [math]n\ge 1[/math] and we must use the plus sign inside the floor function to arrive at a positive integer. We need

[math]n\le n\sqrt{n^2–4m} < n+1[/math]

Or

[math]1 \le \sqrt{n^2–4m} < 1+\dfrac{1}{n}[/math]

Or

[math]1 \le n^2–4m < 1+\dfrac{2}{n}+\dfrac{1}{n^2}[/math]

If [math]n=2[/math] we find [math]1\le 2^2–4m \le 2[/math], which has no solutions [math]m\in \mathbb{N}[/math]. If [math]n>2[/math] then we necessarily have

[math]n^2–4m=1[/math]

Set [math]n=2k-1[/math] and find [math]4k^2-4k=4m[/math], thus [math]m=k(k-1)[/math]. And we find

[math]x=\dfrac{1}{2}(2k-1+1)=k[/math]

[math]y=\dfrac{1}{2}(2k-1–1)=k-1[/math]

Or, if we make sure that [math]x,y>0[/math]:

[math]\boxed{(x,y)\in \{(k+1,k)| \; k\in \mathbb{N} \}}[/math]

Assume without loss of generality that [math]x \geq y[/math]. Then, cubing both sides yields

[math]7x^2 - 13xy + 7y^2 = (x - y + 1)^3. \tag*{}[/math]

Letting [math]t = x - y[/math], then [math]7t^2 = 7(x^2 - 2xy + y^2)[/math]. Then, this can be rewritten as

[math]7t^2 - x \cdot (-(x - t)) = (t + 1)^3. \tag*{}[/math]

Now, we rearrange this equation as a quadratic equation in [math]x[/math]:

[math]x^2 - tx + (-t^3 + 4t^2 - 3t - 1). \tag*{}[/math]

Applying the quadratic formula, solving for [math]x[/math] yields

[math]x = \displaystyle \frac{t \pm \sqrt{(t - 2)^2(4t + 1)}}{2} = \frac{t \pm |t - 2| \sqrt{4t + 1}}{2}. \tag*{}[/math]

Since we want [math]x \in \mathbb{N}[/math] and [math]4t + 1[/math] is an odd integer, this forces [math]4t + 1 = (2k+1)^2[/math] for some non-negative integer [math]k[/math]. Hence [math]t = k^2 + k[/math], and we obtain

[math]x = \displaystyle \frac{(k^2 + k) \pm |k^2 + k - 2| \cdot (2k+1)}{2}. \tag*{}[/math]

Next since we want [math]x > 0[/math], we take the plus sign in the formula for [math]x[/math] above:

[math]x = \displaystyle \frac{(k^2 + k) + |(k+2)(k-1)| \cdot (2k+1)}{2}. \tag*{}[/math]

Hence, solving for [math]y[/math] yields

[math]\displaystyle y = x - t = x - (k^2 + k) = \frac{-(k^2 + k) + |(k+2)(k-1)| \cdot (2k+1)}{2}. \tag*{}[/math]

If [math]k = 0[/math], then we obtain [math](x, y) = (1, 1)[/math]. If [math]k = 1[/math], then [math]y < 0[/math] and we thus ignore this solution.

Hence, we can assume that [math]k \geq 2[/math] and remove the absolute value signs. This yields [math]x = k^3 + 2k^2 - k - 1[/math]; this is easily checked to always be positive. Similarly, [math]y = k^3 + k^2 - 2k - 1 > 0[/math] for these values of [math]k[/math].

Therefore, all positive integers to the given Diophantine equation are given by

[math]\boxed{(x, y) = (1, 1), \, (k^3 + 2k^2 - k - 1, k^3 + k^2 - 2k - 1) \text{ where } k \in \mathbb{N}_{\geq 2}.} \tag*{}[/math]

[math]\boxed{[x,y]=[6,3]}[/math] is the only solution.

[math]y[/math] being odd, [math]2^x \equiv 0 \pmod 4[/math], so [math]x \ge 2[/math]

If [math]x[/math] is even, [math]2^x[/math] is a perfect square

If [math](2^\frac{x}{2}+1)^2-2^x=2^{\frac{x}{2}+1}+1>17[/math] or [math]x>6[/math], [math]2^x+17[/math] can never be a perfect square.

But, instead of checking for [math]x=[2,4,6][/math], we see that [math]2^6+17=81[/math], so [math]y \le 3[/math]

[math]y=3[/math], [math]x=6[/math] works. For [math]y \le 2[/math], [math]2^x+17>y^4[/math], so we don’t have any other solution.

If [math]x[/math] is odd, [math]2^x+17[/math] ends in [math][5,9][/math]. No fourth power ends in [math]9[/math]. So, [math]2^x+17[/math] ends in [math]5[/math] and [math]2^x[/math] ends in [math]8[/math], which means that [math]x=4m-1[/math], [math]m \ge 1[/math], and [math]y \equiv 0 \pmod 5[/math]

[math]2^{4m-1}+17=y^4 \implies 2^{4m}=2(y^4-17)[/math]

[math]y^4 \equiv [-4,-1,0,1,4] \pmod {17}[/math], so, [math]2(y^4-17) \equiv [-8,-2,0,2, 8] \pmod {17}[/math], but the [math]LHS=2^{4m} \equiv [-1,1] \pmod {17}[/math], so no solutions in odd [math]x[/math] exist.

We want to solve the Diophantine equation

[math]x^2 - 12xy - 4x + 11y^2 - 6y - 5 = 0. \tag*{}[/math]

First, we complete the square with the first two terms in [math]x[/math]:

[math](x - 6y)^2 - 4x - 25y^2 - 6y - 5 = 0. \tag*{}[/math]

Next, we complete the square in [math]x - 6y[/math]:

[math]((x - 6y)^2 - 4(x - 6y) + 4) - 25y^2 - 30y - 9 = 0 \tag*{}[/math]

so that

[math](x - 6y - 2)^2 - 25y^2 - 30y - 9 = 0. \tag*{}[/math]

Finally, we complete the square in [math]y[/math]:

[math](x - 6y - 2)^2 - (5y + 3)^2 = 0. \tag*{}[/math]

As a result of our algebraic manipulations above, we can now readily factor the equation above via difference of two squares:

[math]((x - 6y - 2) + (5y + 3))((x - 6y - 2) - (5y + 3)) = 0, \tag*{}[/math]

and therefore

[math](x - y + 1)(x - 11y - 5) = 0. \tag*{}[/math]

Hence, it follows that [math]x = y - 1[/math] or [math]x = 11y + 5[/math], and this yields the solutions in the integers

[math]\boxed{(x, y) = (t-1, t) \text{ or } (11t + 5, t), \text{ where } t \in \mathbb{Z}}. \tag*{}[/math]

Depending on whether we want non-negative integer or positive integer solutions, we can restrict [math]t[/math] accordingly.

How do I find all pairs of real numbers [math](x,y),[/math] which satisfy the following system of equations:
[math]x^6=y^4+18[/math] and [math]y^6=x^4+18?[/math]

Subtracting the second equation from the first equation, we get,

[math]x^6-y^6+x^4-y^4=0[/math]

[math]\Rightarrow (x^2-y^2)(x^4+x^2y^2+y^4)+(x^2-y^2)(x^2+y^2)=0.[/math]

[math]\Rightarrow (x^2-y^2)(x^4+x^2y^2+y^4+x^2+y^2)=0.[/math]

[math]\Rightarrow x^2-y^2=0[/math] or [math]x^4+x^2y^2+y^4+x^2+y^2=0.[/math]

If [math]x^4+x^2y^2+y^4+x^2+y^2=0[/math] then [math](x,y)=(0,0),[/math] which does not satisfy the given equations.

[math]\Rightarrow x^2-y^2=0.[/math]

[math]\Rightarrow x^2=y^2.[/math]

Substituting this in the first equation, we get, [math]x^6-x^4-18=0.[/math]

[math]\Rightarrow x^6-3x^4+2x^4-6x^2+6x^2-18=0.[/math]

[math]\Rightarrow (x^2-3)(x^4+2x^2+6)=0.[/math]

[math]\Rightarrow x^2-3=0[/math] or [math]x^4+2x^2+6=0.[/math]

The second equation does not have any solution in the set of real numbers.

[math]\Rightarrow x=\pm\sqrt 3[/math] and consequently [math]y=\pm\sqrt 3.[/math]

[math]\Rightarrow[/math] The real valued solutions of the pair of equations are [math](x,y)=(\sqrt 3,\sqrt 3)[/math] or [math](\sqrt 3,-\sqrt 3)[/math] or [math](-\sqrt 3,\sqrt 3)[/math] or [math](-\sqrt 3,-\sqrt 3).[/math]

*A2A

Trusting my instincts here and taking a blind shot. Not even sure if this is a good way to solve.

Considering the function [math]y=x^2[/math], or [math]x=y^2[/math]. These are parabolas with a [math]x[/math] and [math]y[/math] symmetry respectively.

So, if we are considering the equation [math]y^2=x^4[/math], or [math]x^2=y^4[/math], they represent a pair of parabolas. The given equations represents a pair of parabolas.

From the first equation:

[math]x^2+x=0 \\ \implies x(x+1)=0\\ \implies x=-1,0\tag*{}[/math]

[math]y^4+y^3+y^2+y=0\\ \implies y^3(y+1)+y(y+1)=0\\ \implies (y+1)(y^3+y)=0\\ \implies y(y+1)^2=0\\ \implies y=-1,-1,0\tag*{}[/math]

Making a combination:

[math](x,y)=\{(0,0),(0,-1),(-1,0),(-1,-1)\}\tag*{}[/math]

This actually works because these pairs of values can actually satisfy both sides of the given equations. I admit its not the best way to solve an equation.

Fundamental Theorem of Algebra: An [math]n^{\text{th}}[/math] order polynomial has exactly [math]n[/math] roots.

I think I got away with this one. These should be the only solutions, according to the theorem.

[math]\\[/math]

[math]x^2y^2 − 2x^2y + 3x^2 + 4xy + 2y^2 = 4x + 4y + 1 \tag {E01}[/math]

[math]\\[/math]

[math](x^2 + 2) \cdot y^2 − (2x^2 - 4x + 4) \cdot y + (3x^2 - 4x - 1) = 0 [/math]

[math]\text{Above is a quadratic equation in 'y'. For } \: x,y \in \mathbb{Z} : \Delta \ge 0[/math]

[math]\implies (2x^2 - 4x + 4)^2 - 4(x^2+2) \cdot (3x^2 - 4x - 1) \ge 0 [/math]

[math]\implies x^4 - 4x^3 + 8x^2 - 8x + 4 -3x^4 + 4x^3 + x^2 - 6x^2 + 8x + 2 \ge 0 [/math]

[math]\implies 2x^4 - 3x^2 - 6 \le 0 \implies 16x^4 - 24x^2 - 48 \le 0 [/math]

[math]\implies \left(4x^2 - 3 \right)^2 \le 57 [/math]

[math]\implies \dfrac{3 - \sqrt{57}}{4} \le x^2 \le \dfrac{3 + \sqrt{57}}{4} [/math]

[math]\implies x \in \left[\: \dfrac{- \sqrt{3+\sqrt{57}}}{2} \:,\: \dfrac{\sqrt{3+\sqrt{57}}}{2}\right] [/math]

[math]\implies x \in [\: -1.62 \:,\: 1.62 ] \implies x \in [-1, 1] \quad \text{Considering only } \mathbb{Z} \: \text{ values} \\[/math]

[math]\text{Case - 1 : When x = -1 : }[/math]

[math]\qquad (E01) : \: (-1)^2y^2 − 2(-1)^2y + 3(-1)^2 + 4(-1)y + 2y^2 = 4(-1) + 4y + 1 [/math]

[math]\qquad \implies 3y^2 − 10y + 6 = 0 [/math]

[math]\qquad \implies y = \dfrac{10 \pm \sqrt{100 - 4(3)(6)}}{6} = \dfrac{5 \pm \sqrt{7}}{3} \: \notin \mathbb{Z} \: \times \\[/math]

[math]\text{Case - 2 : When x = 0 : } [/math]

[math]\qquad (E01) : \: 2y^2 = 4y + 1 \implies 2y^2 - 4y - 1 = 0 [/math]

[math]\qquad \implies y = \dfrac{4 \pm \sqrt{16 + 4(1)(2)}}{4} = \dfrac{2 \pm \sqrt{6}}{2} \: \notin \mathbb{Z} \: \times \\[/math]

[math]\text{Case - 3 : When x = 1 : } [/math]

[math]\qquad (E01) : \: (1)^2y^2 − 2(1)^2y + 3(1)^2 + 4(1)y + 2y^2 = 4(1) + 4y + 1 [/math]

[math]\qquad \implies 3y^2 - 2y - 2 = 0 [/math]

[math]\qquad \implies y = \dfrac{2 \pm \sqrt{4 + 4(3)(2)}}{6} = \dfrac{1 \pm \sqrt{7}}{3} \: \notin \mathbb{Z} \: \times \\[/math]

[math]\qquad \therefore \: \boxed{\: \mathbf{\text{No integer solution exists.}}\:}[/math]

We want to find all positive integer solutions to the Diophantine equation

[math]2^x + 17 = y^4. \tag*{}[/math]

In order to accomplish this, we first show that [math]x[/math] must be even. We show this by reducing the given equation modulo [math]17[/math]:

[math]2^x \equiv y^4 \bmod 17. \tag*{}[/math]

Then, raising both sides to the fourth power yields

[math]2^{4x} \equiv y^{16} \bmod 17. \tag*{}[/math]

The right side of the congruence either is [math]1 \bmod 17[/math] when [math]17 \nmid y[/math] (by Fermat’s Little Theorem), or [math]0 \bmod 17[/math] when [math]17 \mid y[/math]. The latter case cannot occur, because there is no solution to [math]2^n \equiv 0 \bmod 17[/math]. Hence [math]17 \nmid y[/math], and we have

[math]16^x \equiv 2^{4x} \equiv 1 \bmod 17. \tag*{}[/math]

Since [math]16^x \equiv (-1)^x \bmod 17[/math], it is easily checked that [math]x[/math] must be even.

Now that we know [math]x[/math] is even, we can rearrange the given Diophantine equation and factor via difference of squares, giving us

[math]y^2 - 2^x = (y^2 + 2^{x/2})(y^2 - 2^{x/2}) = 17. \tag*{}[/math]

Since [math]17[/math] is prime and [math]y^2 + 2^{x/2} > y^2 - 2^{x/2}[/math], we must have

[math]y^2 + 2^{x/2} = 17 \text{ and } y^2 - 2^{x/2} = 1. \tag*{}[/math]

Adding these equations together, we obtain [math]2y^2 = 18[/math]. Since [math]y > 0[/math], this yields [math]y = 3[/math]. Finally, substituting this into either equation above, we see that [math]x = 6[/math].

Hence, the only solution in positive integers to the given Diophantine equation is

[math]\boxed{(x, y) = (6, 3)}. \tag*{}[/math]

[math]\boxed{[x,y]=[[1,1], [1,2], [2,1], [2,5], [3,2], [3,5]]}[/math] is the set of solutions.

We need [math]xy \mid (x^2+1)(y+1) [/math]

[math]\implies x^2+y+1=kxy[/math], [math]k \ge 1[/math] (ignoring [math]x^2y[/math] because [math]xy \mid x^2y[/math])

(i) [math]x=y \implies x^2+x+1=kx^2[/math]

[math]\implies k=1+\frac{x+1}{x^2}[/math], which is an integer [math]k=3[/math] only for [math]x=1[/math]. So, [math]x=y=1[/math] is a verifiable solution (a rather trivial one).

(ii) [math]x>y[/math]; Here, in [math]x(ky-x)=y+1[/math], the [math]RHS>0[/math], so [math]ky-x \ge 1[/math]; but because [math]x>y[/math], [math]x=y+1[/math]. If [math]x>y+1[/math], the [math]RHS<LHS[/math].

This means that [math]ky-y-1=1 \implies k=1+\frac{2}{y}[/math], which has integer values for [math]y=[1,2][/math] as [math]k=[3,2][/math]. Correspondingly, [math]x=[2,3][/math], so [math][x,y]=[2,1][/math] and [math][3,2][/math] are solutions.

(iii) [math]x<y[/math]; [math](kx-1)y=x^2+1 \implies kx-1 \le x[/math] because [math]xy \ge x^2+1[/math] for [math]y \ge 2[/math] (with the equality applicable only when [math][x,y]=[1,2][/math])

[math]\implies k \le 1+\frac{1}{x}[/math] (1), which means that [math][k,x]=[2,1][/math] is a set of parameters worth checking for. Although the above equality discussion spells the solution out, we could check with a substitution and see that [math]y+2=2y \implies y=2[/math] and, so, [math][x,y]=[1,2][/math] is another solution.

The last solutions relate to [math]k=1[/math] based on (1), which means that:

[math]y=\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}[/math]

[math]x=[2,3][/math] work to generate integer [math]y=[5,5][/math], so [math][x,y]=[2,5][/math] and [math][3,5][/math] are the last solutions.

[math]\\[/math]

[math]xy \cdot (x^2y^2 − 12xy − 12x − 12y + 2) = (2x + 2y)^2 \tag {E01}[/math]

[math]\implies xy \cdot ( x^2y^2 - 12xy + 2) + 9x^2y^2 = (2x + 2y)^2 + 6xy \cdot (2x+2y)+ 9x^2y^2[/math]

[math]\implies xy \cdot \left[ x^2y^2 - 3xy + 2 \right] = (2x + 2y + 3xy)^2 [/math]

[math]\implies xy \cdot (xy - 1) \cdot (xy - 2) = (2x + 2y + 3xy)^2 \tag {E02}[/math]

[math]\\[/math]

[math]\text{Trivial Solutions : LHS = 0 When : }[/math]

[math]\qquad \text{Case T01 : When x = 0 : } [/math]

[math]\qquad \qquad (E02) : \: (2x + 2y + 3xy) = 0 [/math]

[math]\qquad \qquad \implies y = 0 \implies \boxed{\: (x,y) = (0,0) \:} \\[/math]

[math]\qquad \text{Case T02 : When y = 0 : } [/math]

[math]\qquad \qquad (E02) : \: (2x + 2y + 3xy) = 0 [/math]

[math]\qquad \qquad \implies x = 0 \implies \boxed{\: (x,y) = (0,0) \:} \\[/math]

[math]\qquad \text{Case T03 : When xy = 1 : } [/math]

[math]\qquad \qquad (E02) : \: (2x + 2y + 3xy) = 0 [/math]

[math]\qquad \qquad \implies (x+y) = \frac{-3}{2} \:; \quad xy = 1 [/math]

[math]\qquad \qquad x,y \text{ are roots of : } \: t^2 + \frac{3t}{2} + 1 = 0[/math]

[math]\qquad \qquad \implies x,y = t = \dfrac{\frac{-3}{2} \pm \sqrt{\frac{9}{4} - 4}}{2} \: \notin \mathbb{Z} \: \times \\[/math]

[math]\qquad \text{Case T04 : When xy = 2 : } [/math]

[math]\qquad \qquad (E02) : \: (2x + 2y + 3xy) = 0 [/math]

[math]\qquad \qquad \implies (x+y) = -3 \:; \quad xy = 2 [/math]

[math]\qquad \qquad x,y \text{ are roots of : } \: t^2 + 3t + 2 = 0[/math]

[math]\qquad \qquad \implies (t+1) \cdot (t+2) = 0 [/math]

[math]\qquad \qquad \boxed{\: x,y = t = \left\{\: (-1 \:,\: -2 ) \:,\: (-2, -1) \: \right\} \:}\\[/math]

[math]\text{For other solutions, let : (x+y) = axy } [/math]

[math]\qquad (E02) : \: xy \cdot (xy - 1) \cdot (xy - 2) = (axy + 3xy)^2 [/math]

[math]\qquad \implies (xy - 1) \cdot (xy - 2) = xy \cdot (a+3)^2 \tag {E03}[/math]

[math]\qquad \text{Since x and y are integers AND } [/math]

[math]\qquad \qquad \text{xy and (xy-1) are relatively co-prime } [/math]

[math]\qquad \qquad \text{(xy-1) and (xy-2) are relatively co-prime } [/math]

[math]\qquad \qquad \text{RHS has a square term as one of the factors } [/math]

[math]\qquad \text{No integer solution is possible.} [/math]

Letting [math]x = y + t[/math] for some integer [math]t[/math], we obtain

[math](y+t)^3 + 3(y+t)^2 + 2(y+t) + 9 = y^3 + 2y, \tag*{}[/math]

and thus

[math](3t + 3)y^2 + (3t^2 + 6t)y + (t^3 + 3t^2 + 2t + 9) = 0. \tag*{}[/math]

If [math]t = -1[/math], then [math]y = 3[/math]; this gives us the solution [math](x, y) = (2, 3)[/math].

Otherwise assuming [math]t \neq -1[/math] gives us a quadratic equation in [math]y[/math], and this can only possibly have integer solutions if its discriminant is nonnegative. Computing the discriminant, we obtain

[math]\begin{align*} \Delta &= (3t^2 + 6t)^2 - 4(3t + 3)(t^3 + 3t^2 + 2t + 9)\\ &= -3 (t^4 + 4t^3 + 8t^2 + 44t + 36). \end{align*} \tag*{}[/math]

Clearly, [math]\Delta < 0[/math] when [math]t \geq 0[/math]. Moreover, [math]\Delta < 0[/math] when [math]t \leq -5[/math], because

[math]\begin{align*} \Delta &= t^2 (t + 2)^2 + (2t + 11)^2 - 85\\ &\geq (-5)^2 \cdot (-5 + 2)^2 + (2 \cdot (-5) + 11)^2 - 85\\ &> 0. \end{align*} \tag*{}[/math]

(The penultimate line is best seen from minimizing each square quantity separately).

Then, only values of [math]t[/math] that remain to be checked are

[math]t \in \{-4, -3, -2\}. \tag*{}[/math]

(In each of these cases, [math]\Delta \geq 0[/math]). However, we also need [math]\Delta[/math] to be a perfect square to ensure that [math]y[/math] is at least rational. This only occurs when [math]t = -4[/math], and this yields the integer solution [math](x, y) = (-3, 1)[/math]. However, we ignore this solution, because we only want positive integer solutions.

Therefore, we conclude that the only positive integer solution to the given Diophantine equation is

[math]\boxed{(x, y) = (2,3)}. \tag*{}[/math]

Essentially, we want to find all integers [math]x[/math] such that

[math]x^{12} + x^6 + 1 \equiv 0 \bmod 19. \tag*{}[/math]

By using the difference of cubes formula, we start by completing the square with the middle term:

[math]\displaystyle \frac{x^{18} - 1}{x^6 - 1} \equiv 0 \bmod 19. \tag*{}[/math]

Since [math]19[/math] is prime, Fermat’s Little Theorem, any [math]x \in \{1, 2, 3, …, 18\} \bmod 19[/math] makes the numerator equal to [math]0 \bmod 19[/math]. Hence, we only need to exclude the solutions to [math]x^6 - 1 \equiv 0 \bmod 19[/math]. Factoring yields

[math]\begin{align*} x^6 - 1 &\equiv (x^3 - 1)(x^3 + 1)\\ &\equiv (x - 1)(x^2 + x + 1) \cdot (x + 1)(x^2 - x + 1)\\ &\equiv (x - 1)(x^2 - 18x + 77) \cdot (x + 1)(x^2 - 20x + 96)\\ &\equiv (x - 1)(x - 7)(x - 11) \cdot (x + 1)(x - 8)(x - 12) \bmod 19. \end{align*} \tag*{}[/math]

Solving this (for positive residues) yields [math]x \equiv 1, 7, 11, 18, 8, 12 \bmod 19[/math].

Hence, we conclude that the integers [math]x[/math] satisfying the given Diophantine equation are given by

[math]\boxed{ x \equiv 2, 3, 4, 5, 6, 9, 10, 13, 14, 15, 16, 17 \bmod 19}. \tag*{}[/math]

[math]\\[/math]

[math]\text{Let } \: k = \dfrac{x^2y + x + y}{xy^2 + y + 11} \tag {E01} \\[/math]

[math]\text{Trivial Case - 1 : When x = 0 : } [/math]

[math]\qquad k = \dfrac{y}{y + 11} \implies ky + 11k = y \implies y = \dfrac{11k}{1 - k} [/math]

[math]\qquad \text{Possible integer solutions : } k = \left\{\: -10 \:,\: 0\:,\: 2 \:,\: 12 \:\right\} [/math]

[math]\qquad \implies \boxed{\: (x,y) = \left\{\: (\:0 \:,\: -10\:) \:,\: (\:0 \:,\: 0\:) \:,\: (\:0 \:,\: -22\:) \:,\: (\:0 \:,\: -12\:) \:\right\}\:} \\[/math]

[math]\text{Trivial Case - 2 : When y = 0 : } [/math]

[math]\qquad k = \dfrac{x}{11} \implies x = 11n[/math]

[math]\qquad \implies \boxed{\: (x,y) = (\:11n \:,\: 0\:) \:\forall \: n \: \in \: \mathbb{Z} \:} \\[/math]

[math]\text{Trivial Case - 3 : When x = y : } [/math]

[math]\qquad k = \dfrac{x^3 + 2x}{x^3 + x + 11} [/math]

[math]\qquad \implies (k-1) = \dfrac{x - 11}{x^3 + x + 11} [/math]

[math]\qquad \implies \boxed{\: (x,y) = (\:11 \:,\: 11\:) \:} \\[/math]

I haven’t worked out the remaining part of generic solutions.

[math]\\[/math]

[math]\text{Given : } \: x,y \in \mathbb{Z^+} [/math]

[math]x^3 + y^3 = x^2 + 42xy + y^2 \tag {E01}[/math]

[math]\text{Trivial Solution : When x = y : } [/math]

[math](E01) :\: x^3 + x^3 = x^2 + 42x^2 + x^2 \implies 2x^3 = 44x^2 [/math]

[math]\implies \boxed{\: x = y = 22 \:} \quad \left[\: \because \: x ,y > 0 \: \right] [/math]

[math](E01) : \: (x+y) \cdot (x^2 - xy + y^2) = (x^2 - xy + y^2) + 43xy [/math]

[math]\implies (x+y-1) \cdot (x^2 - xy + y^2) = 43xy \tag {E02}[/math]

[math]\\[/math]

[math]\text{Possible Solutions : } [/math]

[math]\text{Case - 1 : When }\: (x+y-1) = 43; \quad (x^2-xy+y^2) = xy \: : [/math]

[math]\qquad \implies x=y; \quad x+y = 44 \implies \boxed{\: x = y = 22\:} \\[/math]

[math]\text{Case - 2 : When }\: (x+y-1) = x; \quad (x^2-xy+y^2) = 43y \: : [/math]

[math]\qquad \implies y=1; \implies x^2 - x + 1 = 43 [/math]

[math]\qquad \implies x^2 - x - 42 = 0 \implies (x-7)(x+6) = 0 [/math]

[math]\qquad \implies x = 7 \quad \left[\text{ ignoring negative value of 'x' } \right] [/math]

[math]\qquad \implies \boxed{\: (x,y) = (7,1) \: } \\[/math]

[math]\text{Case - 3 : When }\: (x+y-1) = y; \quad (x^2-xy+y^2) = 43x \: : [/math]

[math]\qquad \implies x=1; \implies y^2 - y + 1 = 43 [/math]

[math]\qquad \implies y = 7 \quad \left[\text{ ignoring negative value of 'y' } \right] [/math]

[math]\qquad \implies \boxed{\: (x,y) = (1,7) \: } \\[/math]

[math]\text{Case - 4 : When }\: (x+y-1) = 1; \quad (x^2-xy+y^2) = 43xy \implies y=(2-x) [/math]

[math]\qquad \implies x^2 + 44xy + y^2 = 0 [/math]

[math]\qquad \implies 21x^2 - 42x - 2 = 0 \implies x\notin \mathbb{Z}[/math]

One can try with other possible cases such as : (x+y-1) = 43k and so on. But, there is no other positive integer solution.

Graph from Desmos | Let's learn together. shows that integral values of x and y are bound in the range (0,23). Only integer solutions are : (x,y) = { (1,7), (7,1), (22, 22) }