If $p \equiv 1$ mod 4 is prime then there are 8 pairs $(a,b)$ s.t. $a^2+b^2 = p$

Asked Mar 03 '20 at 14:21

Active Mar 03 '20 at 14:21

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I need help to show the above statement. I know that there is definitely a solution $(a,b)$ and I can take negatives to make 8 different pairs; but how do I show there are no more than these 8 pairs?

Also I need to prove the more general statement that there are $4(k+1)$ pairs s.t. $a^2+b^2 =p^k$

Thanks for your help and hints.

asked Mar 03 '20 at 14:21

citizenfour

You need to know the basic facts about the ring of Gaussian integers. Search for Gaussian integers for details. – WhatsUp Mar 03 '20 at 14:25
1

Here is a nice video by 3blue1brown, where this result is explained and used, and here is a mathologer video doing the exact same thing. – Arthur Mar 03 '20 at 14:29
You can find out the details about Gaussian integers here: https://kconrad.math.uconn.edu/math5230f12/handouts/Zinotes.pdf
Use it to prove the above statement.
– SARTHAK GUPTA Mar 03 '20 at 14:58
Thanks for all your help – citizenfour Mar 03 '20 at 15:19

If $p \equiv 1$ mod 4 is prime then there are 8 pairs $(a,b)$ s.t. $a^2+b^2 = p$

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