<s>[INST] If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed
 [/INST] From what is given, $a,b,c$ are not multiples of $7$. From $(\pm1)^2\equiv 1\pmod 7$, $(\pm2)^2\equiv 4\pmod 7$, $(\pm3)^2\equiv 2\pmod 7$., we see that the only few values for $a^2,b^2,c^2\pmod7$ are possible and the only combinations leading to $a^2+b^2=c^2$ are $1+1\equiv 2$ and $2+2\equiv4$ and $4+4\equiv1$
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