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Show that there are no integers $x, y$ such that $$x^{2015} - y^{2016} = 2115$$

This is a problem in my school competition. The only thing I can think of is considering two sides of the equation in some appropriate modulo to show some contradiction (like 2, 3, 5, 7, 11, 13 etc) but the only thing I've got by far is $y$ must be even and $x=8k+3$

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    $\begingroup$ What have you tried ? What is the context of this question ?. There are methods to show that $3^{2015} \gt 2^{2016}$. Since that is the case, no other integers will work. $\endgroup$ – Shailesh Sep 13 '15 at 7:05
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    $\begingroup$ @Shailesh: Care to elaborate on that? I fail to see how non-existence of $(x,y)$ follows from the fact that $3^{2015}>2^{2016}$. $\endgroup$ – Jyrki Lahtonen Sep 13 '15 at 7:13
  • $\begingroup$ Primitiveroot: Give some context, please? This smells a bit like a contest problem. Nothing bad about that, but if it is, where did you find it? What approaches have you tested and eliminated? $\endgroup$ – Jyrki Lahtonen Sep 13 '15 at 7:21
  • $\begingroup$ @JyrkiLahtonen Sure Jyrki. Unless I have missed something, and I am always willing to correct myself and learn, (1) if the integers are not consecutive the gap is wider (2) Even for consecutive integers x, y; $x^{2015} - y^{2016}$ will go on increasing as x increases. $\endgroup$ – Shailesh Sep 13 '15 at 7:24
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    $\begingroup$ @Shailesh: Hmm. But the gap can be zero. For example when $x=2^{2016}$ and $y=2^{2015}$. Note that in that case $x=2y$. No need for $x$ and $y$ to be anywhere near consecutive for the powers to be close to each other. $\endgroup$ – Jyrki Lahtonen Sep 13 '15 at 7:25
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Well, 31 is the lowest modulo in which this thing is not solvable. $x^{2015}\mod31$ can be either of (0, 1, 5, 6, 25, 26, 30), $y^{2016}\mod31$ is from (0, 1, 2, 4, 8, 16), the difference has to be 7, and it is just not there.

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    $\begingroup$ Nice! Too bad we need to wait for 30 years to reuse this property of $\Bbb{Z}_{31}^*$ as a contest problem. $\endgroup$ – Jyrki Lahtonen Sep 13 '15 at 15:03
  • $\begingroup$ That was nice! But I think that maybe there's an alternative solution, because it is pretty difficult to consider $x^{2015}$ or even $x^5$ $mod$ $31$ in the exam room without calculator. $\endgroup$ – primitiveroot Sep 13 '15 at 22:56
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    $\begingroup$ @primitiveroot I think this is what is expected. You should be clued in to examining $31$ as the largest prime divisor of of $2015$. The multiplicative group mod $31$ is cyclic and of order $30$. Since $30$ is divisible by both $5$ and $6$, you know that $5$th and $6$th powers will be restricted (having $6+1$ and $5+1$ possible values respectively). Since $5\mid2015$ and $6\mid2016$, you have the difference of a $5$th power and a $6$th power. It's easy to calculate and list the $6+1$ and $5+1$ values as Ivan has done, without using a calculator. Just do the modular arithmetic in steps. $\endgroup$ – alex.jordan Sep 20 '15 at 17:35

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