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This is a question I am attempting. We are asked to use Fermat's Little Theorem to show that $x^2 = 7 + 13y^2$ has no integer solutions.

My attempt: I chose to work in modulo 13. Using Fermat's Little Theorem, I get $x^{12}=1(mod13)$ and from the question I got $x^2=7(mod13)$ and $7=-6(mod13)$. I think I need to get somewhere such that I get $x^2=a(mod13)$ for some $a$ and then notice how this has no solutions but I'm neither sure how to do this nor whether this would this imply there are no integer solutions for both $x$ and $y$?

I'm clearly struggling with this proof and would appreciate any directions.

Thanks

Francis
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  • Seem like a good argument to me. Any integer solution with $n^2=7+13m^2$ would give us a square root of $7\pmod {13}$, namely $n$. But you say you have shown that $7$ is not a square $\pmod {13}$ so this would be a contradiction. – lulu Oct 31 '18 at 17:45
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    Raise $,x^2\equiv 7,$ to the $6$th power to get $1\equiv -1$ or, equivalently, apply Euler's criterion. – Bill Dubuque Oct 31 '18 at 17:49
  • Your way is fine. Just to note that another way of doing this is modulo $7$ when you get $x^2\equiv -y^2$. Clearly if $x$ and $y$ are both divisible by $7$ they give no solution to the original equation. $y$ therefore has a multiplicative inverse modulo $7$ and $-1$ would have to be a square mod $7$ - but this is not the case. There are ways of making this argument more efficient. – Mark Bennet Oct 31 '18 at 18:01

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Your approach is good. As you have noted, working modulo $13$ shows that if a solution exists then $$x^2\equiv7\pmod{13}.$$ This implies that $7^6\equiv x^{12}\equiv1\pmod{13}$ by Fermat's Little Theorem. Now it suffices to check that $$7^6\not\equiv1\pmod{13},$$ to reach a contradiction. This allows you to conclude that no solution exists.

Servaes
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