Do you know how to do this proof?
Suppose (a , b)=1 . Prove that if p|(a^2+b^2) then p ≡ 1 (mod 4).
Note: p is any odd prime.
Thus, the Legendre Symbol (I presume you learned them by now),
Since -b^2 is a quadradic residue of a^2 we have,
That is a famous case and only happens when,
p≡1 (mod 4)
This looks greatly similar to Fermat's theorem about expressing 4k+1 primes as a sum of two unique squares.
The proof of that is much more complicated (I wonder how Fermat proved it )
Would do it like this.
Divide the prove into cases. (Divison Algorithm)
'a' can have the form 4k,4k+1,4k+2,4k+3
'a^2' can only be 4k,4k+1
'b' can have the form 4k,4k+1,4k+2,4k+3
The different possible sums between 'a^2+b^2' are,
Since 'p' divides them and is odd it cannot be 4k,4k+2
Thus, the only possible form of 'a^2+b^2' is 4k+1.
The prime 'p' can have two forms 4k+1,4k+3
Now, a number of the form 4k+3 cannot divide a number of the form 4k+1 because they are different forms.