# Solutions for $615+2^x=y^2$ on the integers

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.

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.

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

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.

• Oh, this is even better than working mod $8$, as I suggested. How embarrassing! – Barry Cipra Aug 20 at 19:32

Hint: $$615\not\equiv1$$ mod $$8$$, so we must have $$x\lt3$$.

• Doctor Who has posted an answer with a much better idea than mine. – Barry Cipra Aug 20 at 19:33

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.)