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Determine all integers $x,y$ such that $f:=(3^x-y^3)(x^3-3^y)$ is a perfect square.

What I have thought: the problem seems a bit hard to begin with. I first tried the situation that $\gcd(x,3)=\gcd(y,3)=1$ and was stuck. It would help a lot if we can make some progress in $\gcd(3^x-y^3,x^3-3^y)$, but I failed. However, if you just expand it, it still seems useless.

Any idea which makes progress is welcomed.

Juggler
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  • x=y lol......... – Saketh Malyala Aug 10 '17 at 00:43
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    Here I wrote a program to find all $(x,y)\in[0,1000]^2$, and got [[0, 1], [1, 0], [3, 3], [6, 9], [9, 6], [9, 27], [12, 81], [15, 243], [18, 729], [27, 9], [81, 12], [243, 15], [729, 18]]. From this, it seems like looking at $(3k,3^k)$ and $(3^k,3k)$ may be useful. Unfortunately, showing all solutions are of this form is likely more difficult. – Mark Schultz-Wu Aug 10 '17 at 01:22
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    @SakethMalyala, this would lead to $f(x,y) = -n^2$, which is almost a perfect square, but alas not quite – mdave16 Aug 10 '17 at 01:25
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    @Mark, strangely enough, all the values are also the solutions to $f(x,y) = 0$. – mdave16 Aug 10 '17 at 01:42
  • @GJon, where did you find this problem? $\LaTeX$ tip, \gcd gives a better look cf. $\gcd$ and $gcd$ – mdave16 Aug 10 '17 at 01:46
  • @mdave16 One of my classmates asked me. – Juggler Aug 10 '17 at 01:49
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    The expression can only be equal to one of ${0,2,4,6,7}$ modulo $8$, so it can only be a square when it's divisible by $8$ (and hence by $16$). – Mastrem Aug 10 '17 at 21:28

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