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I've used Fermat's theorem on the sum of two squares to solve the following problem:

Find all (if any) solutions to $x^2+y^2=2015$

Let's look at $x^2+y^2=p$ (1)

$p$ is a solution iff $p=2$ or $p \equiv 1 \pmod{4}$

If we divide $2015$ into primes we get: $5\times13\times31=2015$

And this is where Fermat's theorem kinda comes in..

It can be deduced from Fermat's theorem on sum of two squares that (1) has a solution iff every prime number $p \equiv 3 \pmod{4}$ appears with an even exponent. (2)

We can clearly see that it doesn't and we are finished.

Thus there are no solutions to this problem.

I feel like my attempt at this question falls short since I don't fully comprehend the proof of Fermat's theorem on sum of two square or the special case (2) that can be deduced from the theorem.

If anyone could give me another approach that be much appreciated.

Any comments on my solution would also be nice

One other approach I was thinking about is that $2015$ is odd and thus $(x,y)$ must one be odd and one even. We know that even squares are even and odd squares are odd and sum of even+odd=odd.

We could thus write the equation on the form $2n+1$ and $2n$ and from here im lost!

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$$2015\equiv3\pmod4$$

For any integer $a, a^2\equiv0,1\pmod4$

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  • $\begingroup$ Do you mind elaborating just a little? $\endgroup$ – einar Jun 20 '17 at 8:11
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    $\begingroup$ This link might help explain @lab bhattacharjee's second line: (math.stackexchange.com/questions/99716/…). This is a relatively well-known fact in mathematics, so I suggest studying the proof to see what techniques it uses. $\endgroup$ – Toby Mak Jun 20 '17 at 8:24
  • $\begingroup$ Following this result a square number can be either $0 \pmod 4, (0+0)$; $1 \pmod 4, (0+1)$ or $(1+0)$; and $2 \pmod 4, (1+1)$, so from these conclusions it is not possible for a square number to be $3 \pmod 4$. $\endgroup$ – Toby Mak Jun 20 '17 at 8:26
  • $\begingroup$ Awesome! I have no idea how I could ever go back to not using math.stack! Wonderful community and you learn so much! Much appreciated $\endgroup$ – einar Jun 20 '17 at 8:51
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Your approach was nice, I complete it. Hope this helps...

Since, 2015 is odd

     Odd + Even = Odd

Lets say x=2m, y=2n+1 ;(m,n) are integers

     x^2 + y^2 = 2015

then, (2m)^2 + (2n+1)^2 = 2015

   4m^2 + 4n^2+4n+1 = 2015

   4(m^2+n^2+n) = 2014

   m^2 + n^2 + n = 1007/2

We can see that left side is the sum of integers but 1007/2 is not.

 Hence, it has no solution.
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