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Find the sum of all positive integers $n$ for which $|2^n + 5^n– 65|$ is a perfect square.

It is easy to check for $n<3$ only $n=2$ works. Looking at $n≥3$, $2^n + 5^n– 65= a^2$ For odd $n$, LHS has unit digit $2$ or $8$ but RHS has unit digit $0,4$ or $6$. Hence $n$ cannot be odd. For even $n$ we can rewrite our expression as follows ($n=2k$): $$(a – 2^k)(a+2^k)=5(5^{2k–1} – 13)$$

But I wasn't able to finish off after this. Both LHS and RHS have the same unit digit ($0$). Through trial and error I found that $n=4$ works, but $n=6,8$ don't. Since the questions asks for the sum as is common in Olympiad number theory it's highly unlikely that any other larger values satisfy. I also tried some alternative ways like looking at the equation modulo $4$ but that didn't help.

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2 Answers 2

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Once you know that $n\geq 3$ must be even, let $ n = 2k$.

Then, apart from finitely many values of $k$,

$$5^{2k} < 5^{2k} + 2^{2k} - 65 < 5^{2k} + 2 \times 5^k + 1. $$

This bounds the expression strictly between 2 consecutive perfect squares, so it cannot be a perfect square.

So, we just need to check those finitely many values, which is left as an exercise to the reader.

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  • $\begingroup$ I think some of the inequalities go the other way here... $\endgroup$
    – Mike
    Jan 19, 2021 at 14:45
  • $\begingroup$ @Mike Which ones? Note the "apart from finitely many values of $k$", esp for small $k$. $\endgroup$
    – Calvin Lin
    Jan 19, 2021 at 14:45
  • $\begingroup$ @Mike valid question but in this case the function $f(x)=2\cdot 5^x-2^{2x}$ increases for $x\ge 1$ so Calvin Lin's point is correct $\endgroup$ Jan 19, 2021 at 14:49
  • $\begingroup$ Yes you are right...You are using the fact that the difference between $5^{2k}$ and the next square is at least $5^{k} >> 2^{2k} \pm 65$. I had read the sentence you wrote wrong, my mistake $\endgroup$
    – Mike
    Jan 19, 2021 at 14:52
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    $\begingroup$ @Mike Yes, I'm using the fact that $ 5 \geq 4$ for the RHS. IE This approach wouldn't have worked out if it was for $ |3^n + 2^n - 65 |$ $\endgroup$
    – Calvin Lin
    Jan 19, 2021 at 14:54
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NOTE: This is not a full answer and is intended to give a boost.

Taking modulo $4$, $$a^2-(0)^k\equiv 0\ (\text{mod}\ 4)$$ $$a\equiv0\ (\text{mod}\ 4)$$ Taking modulo $3$, $$a^2\equiv(-1)^n+(-1)^n-2\equiv 0\ (\text{mod}\ 3)$$ $$a\equiv0\ (\text{mod}\ 3)$$ Therefore, $a$ must be divisible by $12$.

Hope this part helps :)

EDIT: @CalvinLin has a wonderful (intended too I guess), so this will be left as intact, or if I should delete this, tell me.

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  • $\begingroup$ Can you narrow down further, in particular $n$ to a finite set $\endgroup$
    – Mike
    Jan 19, 2021 at 14:42
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    $\begingroup$ Please start off by saying this is a partial answer else people may get confused $\endgroup$ Jan 19, 2021 at 14:44
  • $\begingroup$ @AlbusDumbledore: Done! $\endgroup$ Jan 19, 2021 at 15:03
  • $\begingroup$ ok,see Calvin Lin's complete answer too $\endgroup$ Jan 19, 2021 at 15:04
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    $\begingroup$ It's still useful, if not a complete answer , or even the right direction. $\endgroup$ Jan 20, 2021 at 8:57

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