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seen Apr 11 at 6:16

Dec
5
awarded  Popular Question
Feb
2
accepted show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
Feb
2
comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
I get this but $a^2 \equiv k \pmod n $ and not necessarily $a^2 \equiv 1 \pmod n $ .
Feb
2
comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
Yes. the equation has a solution iff $p=4k+1$
Feb
2
comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
${{(-b^2)}^{2k+1}} \equiv 1 \pmod n \Rightarrow {b}^{2^{2k+1}} \equiv {-1} \pmod n $ which implies $ 1 \equiv {-1} \pmod n $ hence contradiction . It sounds fine to me. Is it good?
Feb
2
comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
This makes sense but we're not supposed to use a quadratic residue result yet. Maybe I should read the proof of this statement and try deriving it.
Feb
2
asked show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
Jan
31
comment Finding primes for which a given number is a perfect square.
Yes, I was looking for a direction and not a solution . I enjoyed the severe reduction and limitations of the possibilities :)
Jan
31
awarded  Supporter
Jan
31
awarded  Scholar
Jan
31
comment Finding primes for which a given number is a perfect square.
Thanks a lot. I got p=3,7 as the (only) two solutions
Jan
31
accepted Finding primes for which a given number is a perfect square.
Jan
31
awarded  Student
Jan
31
awarded  Editor
Jan
31
comment Finding primes for which a given number is a perfect square.
sorry!, edited :)
Jan
31
revised Finding primes for which a given number is a perfect square.
added 161 characters in body; edited tags
Jan
31
asked Finding primes for which a given number is a perfect square.