Pythagorean Triples - $c$ divisible by $7$ Let us say that $(a,b,c)$ is a primitive pythagorean triple. Is it true that $c$ is not divisible by $7$?
I tried to show this and my approach was to see what kind of solutions one can get for
$$
(m^2-n^2)^2+(2mn)^2\equiv 0 \pmod{7}
$$
Do I just have to plug in all combinations mod $7$ and see what I get? Is there another approach?
 A: Parametriz3 $a^2+b^2=c^2$ by $a=u^2-v^2$, $b=2uv$, and $c=u^2+v^2$ with integers $u,v$. Since $7$ is $3$ mod $4$, if $7$ divides $u^2+v^2$, then $7^2$ also divides $u$ and $v$, by factoring in the Gaussian integers. Then $7$ divides the $a$ and $b$.
So, indeed, primitive pythagorean triple $a,b,c$ cannot have $c$ divisible by any prime $p$ congruent to $3$ mod $4$. :)
A: Other answers use the result of the general form of the Pythagorean triples but that's not necessary. The squares modulo $7$ are $0,1,2,4$. So the only way that $a^2+b^2 \equiv 0 \pmod 7$ is that $a^2\equiv b^2 \equiv 0 \pmod 7$, but then $a$, $b$ and $c$ are multiple of $7$, but this can't happen since the triple is primitive.
A: 7 times any primitive Pythagorean triplet (such as (3,4,5) -> (21,28,35)) would be divisible by 7, so no.
A: The primitive Pythagorean triple will be $a=u^2-v^2$, $b=2uv$, $c=u^2+v^2$ for some integers $u,v$.
Now the squares modulo $7$ are $0,1,2,4$ and the only case in which the sum of two squares is $\equiv0\pmod{7}$ is when $u\equiv0\pmod{7}$ and $v\equiv0\pmod{7}$.
But then also $a=u^2-v^2$ and $b=2uv$ are divisible by $7$, contradiction.
