<s>[INST] Show a,b,c congruences when $a^2+b^2=c^2$ i would like to know how to prove that if $a^2+b^2=c^2$ then 
a) $3|a$ or $3|b$ 
b) $5|a$ or $5|b$ or $5|c$ 
c) $4|a$ or $4|b$ 
Also that $a^2+b^2+c^2+1$ can't be divisible by 8
Thank you very much!
 [/INST] (a)
(i)If $3\not\mid ab, a\equiv\pm1\pmod3,b\equiv\pm1\implies a^2+b^2\equiv1+1\equiv-1\pmod 3$ 
but $c\equiv\pm1,0\implies c^2\equiv1,0\pmod 3$
(ii)Alternatively, using this, $a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$ will generate all Pythagorean triples uniquely where m, n, and k are positive integers with m > n, m − n odd, and with m and n co-prime.
So, $ab=k2mn\cdot k(m^2-n^2)$
If $3\mid mn, 3\mid b$ we are done else  $(mn,3)=1$
Using Fermat's Little Theorem, 
$m^2\equiv1\pmod 3$ and $n^2\equiv1\implies 3\mid(m^2-n^2)\implies 3\mid a$
(b)
(i) If $5\not\mid ab, a\equiv\pm1,\pm2\pmod3 \implies a^2\equiv1,4\pmod 5$
Similarly, $b^2\equiv1,4\pmod 5$
Observe that $c\equiv0,\pm1,\pm2\pmod5\implies c^2\equiv0,1,4\pmod 5$
If $a\equiv\pm1,b\equiv\pm1, a^2+b^2\equiv2\pmod5$ which is not congruent to any square $\pmod 5$
If $a\equiv\pm2,b\equiv\pm2, a^2+b^2\equiv8\equiv3\pmod5$ which is not congruent to any square $\pmod 5$
So, either $a\equiv\pm2,b\equiv\pm1$ or $a\equiv\pm1,b\equiv\pm2$
(ii)In the alternate way, $a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$
So, $abc=k^32mn(m^4-n^4)$
If $5|m$ or $5|n$ we are done else $(mn,5)=1$
Using Fermat's Little Theorem, 
$m^4\equiv1\pmod 5$ and $n^4\equiv1\implies 5\mid(m^4-n^4)\implies 5\mid(m^2-n^2)$ or $5\mid(m^2+n^2)$
(c)$a=k(m^2-n^2),b=k(2mn),c=k(m^2+n^2)$
As $m-n$ is odd, $2\mid mn\implies 4\mid b$
The last problem :  Observe that the square of any number is $\equiv0,1,4\pmod 8$
</s>