Can you help me prove this number theory claim:
If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$
Thanks
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3The 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
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2You don't even need $p$ to be prime. It's all red herrings. – Arthur May 19 '17 at 17:48
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@badjohn That's a good point, thanks! – Pablo Mello May 19 '17 at 17:50
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Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
Wuestenfux
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