If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$

Question

Can you help me prove this number theory claim:

Thanks

The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier. — badjohn, May 19 '17 at 17:46

score 2 · Accepted Answer · answered May 19 '17 at 17:49

2

Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.

answered May 19 '17 at 17:49

Wuestenfux

score 0 · Answer 2 · answered May 19 '17 at 17:49

0

$p|a^2+b^2-a^2=b^2$

Especially $p|b$

answered May 19 '17 at 17:49

Marios Gretsas

2 Answers2