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In an attempt to understand and solve this problem, I tried to play with some small finite example, one of which is $$x^4+x=y^2+y$$ Playing with Wolfram-Alpha indicates indeed equations of similar form, where a generic parametric solution (i.e. $x,y$ can both be represented as some function of $t$) does not exist, always have only a finite number of integer solutions.

It seems progress can be made if I am able to understand why there is only a finite number of integer solutions for some specific examples. If anyone has any idea please share with me.

cr001
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    It's an elliptic curve, so has only finitely many integer solutions due to Siegel's theorem. – Angina Seng Aug 10 '20 at 17:39
  • Thank you for the reply, I may need to do some reading on elliptic curve as my knowledge is not enough, but if possible can you tell me whether cases like $x^4+x^2=y^2+y$, which has a generic parametric solution, fail to meet the criteria of Siegel's theorem? – cr001 Aug 10 '20 at 17:46
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    $x^4+x^2=y^2+y$ is the same as $(2x^2+1)^2=(2y+1)^2$ and is not an elliptic curve. – Angina Seng Aug 11 '20 at 02:35
  • I see what happens. Looks like when $y$ is a polynomial function of $x$ the expression can be simplified. But this should be discussion of the original problem so I will stop here. Thank you so much. – cr001 Aug 11 '20 at 02:38

2 Answers2

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If $x^4+x = y^2 + y$,

Then $4(x^4+x) = 4(y^2 + y)$

or $(2y + 1)^2 = 4x^4+4x + 1$

But you need form $a^2x^2 \pm 2abx + b^2$ for infinite solution and the RHS polynomial cannot be converted to that form.

EDIT: Adding further why $4x^4+4x + 1$ polynomial cannot be a square for integer $x \gt 1$. I have taken only positive integer $x$ for the proof, as an example.

If the polynomial is a square,

$ \begin {align} 4x^4+4x + 1 &= n^2, \space n \in \mathbb{Z+} \\ 4x + 1 &= n^2 - 4x^4 \\ 4x + 1 &= (n-2x^2)(n+2x^2) \\ \end {align} $

As ($4x + 1$) is positive for $x \ge 0$, $n - 2x^2 \gt 0$.

Say, $n = a + 2x^2$ where $a \in \mathbb{Z+}$

$4x + 1 = a (4x^2 + a)$

This holds true for $a = 1$ and $x = (0, 1)$. But for $x \ge 2$, LHS will be smaller than the RHS.

Math Lover
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  • Can you elaborate on why $a^2x^2 \pm 2abx + b^2$ is needed for infinite solution. Why it is impossible to be the case that $4x^4+4x+1$ just happens to be a perfect square for infinite many times? – cr001 Aug 10 '20 at 18:14
  • @cr001 I have added some details that may answer your question on why $4x^4+4x+1$ cannot be a perfect square for infinite many times. – Math Lover Aug 10 '20 at 20:58
  • Thank you for clarification. As the other answer also shows, seems the key idea is inequality. After testing with other higher degree examples it feels pretty hard to generalize the idea but I will try. – cr001 Aug 11 '20 at 02:34
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for positive, both bigger than $1$

if $y \geq x^2,$ then $y^2 + y \geq x^4 + x^2 > x^4 + x$

if $1 < y \leq x^2 - 1,$ $y^2 + y \leq x^4 - x^2 < x^4 + x$

I guess there are 3 more cases $x < -1, y > 1,$ then $x > 1, y < -1,$ then $x < -1, y < -1$

Will Jagy
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  • Thank you for the idea of inequality. Seems this approach is possible to gain some insight to the original problem. I will try with other examples. – cr001 Aug 10 '20 at 17:56