Show that if $\gcd(a,b)=1$ the number $a^2+b^2$ hasn't prime factors of the form $4k-1$ and furthemore $\gcd(a,3)=1$ then, the number $a^2+3b^2$ hasn't prime factor of the form $6k-1$.
My approach: For the theorem of sum of two squared, we need that prime divisor has the form $p=1\mod 4$, and therefore $a^2+b^2$ hasn't factor of the form $4k-1$