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This problem is very similar to a popular one, but I found it in this way. I thought it could be solved in a similar manner. This means that $x$ has to be an even number, and then it holds $$615=y^2-2^{2k}=(y-2^k)(y+2^k)$$

possible pair of factors of $615$ are $\{(615,1), (123, 5), (3,205),(15,41)\}$. Then the way this problem is usually solved is by adding the 2 factors and finding the value for for $2^k$. However this time I tried to susbstract the factors so I could find a possible value of $2^k$, but this means we only have the 4 possibilities for the value of $2^k$: $\{614, 118, 2020, 26\}$. Which none are values for $2^k$ with $k\in\Bbb{Z}$. Does this mean there are no integer solutions for this equation? or maybe there's something wrong with my reasoning.

Thanks in advance!

EDIT: I did not assume that $x$ is even, I should have elaborated on that. If $y^2$ is an integer, then the digit on the units place must be one of the followings: $\{1, 4, 5, 6, 9\}$. Powers of 2 can only have the following digits on the units place: $\{2, 4, 6, 8\}$. If $x$ is an odd number, then $2^x$ has either a $2$ or an $8$ as its units place, this in turn means that $y^2=615+2^x$ has either $7$ or $3$ on the units place, which is a contradiction. That's why $x$ must be an even number.

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  • $\begingroup$ See here, or here or this one for the "other" equation $615+y^2=2^x$. $\endgroup$ – Dietrich Burde Aug 20 at 19:13
  • $\begingroup$ @DietrichBurde, those aren't quite the same; the power of $2$ and the square have swapped sides. $\endgroup$ – Barry Cipra Aug 20 at 19:15
  • $\begingroup$ The $2^x$ and $y^2$ are swapped in my post compared to what you just shared. $\endgroup$ – NotAMathematician Aug 20 at 19:16
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    $\begingroup$ Yes, this is a "swapped" version, which is in fact more interesting. $\endgroup$ – Dietrich Burde Aug 20 at 19:17
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    $\begingroup$ @NotAMathematician, very nice proof! $\endgroup$ – Barry Cipra Aug 20 at 21:39
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Suppose $x \geq 2$. Reduce both sides mod 4 to get that $3 \equiv y^2$, a contradiction since $0$ and $1$ are the only squares mod 4.

Then the only possible choices are $x = 0$ and $x = 1$. But neither $615 + 2^0$ nor $615 + 2^1$ is a perfect square. So there are no solutions.

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  • $\begingroup$ Oh, this is even better than working mod $8$, as I suggested. How embarrassing! $\endgroup$ – Barry Cipra Aug 20 at 19:32
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Hint: $615\not\equiv1$ mod $8$, so we must have $x\lt3$.

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  • $\begingroup$ Doctor Who has posted an answer with a much better idea than mine. $\endgroup$ – Barry Cipra Aug 20 at 19:33
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Yes, that prove $615+ 2^{x=2k} = y^2$ has no integer solutions if $x$ is even.

If $x$ is odd we could try.

$615 + 2^{x=2k+1} = y^2$

$2^{2k+1} = y^2 - 615$ so $y$ is odd let $y=2m+1$

$2^{2k+1} = 4m^2 + 4m -614$

$2^{2k} = 2m^2 +2m - 307$ which means $2^{2k}$ is odd so $2^{2k} =1$ and $k =0$

$2m^2 +2m = 308$

$m(m+1) = 154$

But $154 = 2*7*11$ can not be so factored.

So $615+2^x =y^2$ has no integer solutions if $x$ is odd either.

BUt that's pretty inefficient and I don't advise it.

(THis could however give us a hint as to considering arithmetic $\mod 4$ and Doctor Who's answer well eventually fall into place.)

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