If $p \equiv 1 \pmod{4}$, then $p \nmid a^2+3b^2$? Suppose that $a, b$ are co-prime numbers and $p$ is a prime number in the form $4k+1$. Can $3a^2+b^2$ have a prime divisor like $p$?
I really don't know how to think about this! Please help!

Suppose we have $$a^2\equiv -3b^2\pmod p$$
Now, since $\gcd(a,b)=1$ we can exclude the case where one, hence both, of $a,b$ are divisible by $p$.  Thus we can divide by $b^2$ to see that $-3$ must be a square $\pmod p$.
Assume that $p\equiv 1 \pmod 4$.  Quadratic reciprocity then tells us that $-3$ is a square $\pmod p$ if and only if $p\equiv 1\pmod 3$.  Thus we require $p\equiv 1 \pmod {12}$.
It is easy to work backwards.  That is, given $a$ such that $a^2\equiv -3 \pmod p$ then $a^2+3\times 1^2$ is divisible by $p$.
Conclusion:  if $p\equiv 1 \pmod {12}$ then we can find coprime $a,b$ with $a^2+3b^2\equiv 1 \pmod {p}$.
Worth noting:  the same technique can applied to the case $p\equiv -1 \pmod 4$.  Again we require that $=3$ be a square $\pmod p$ but again reciprocity forces us to require that $p\equiv 1 \pmod 3$  Thus we are lead to considering $p\equiv 7 \pmod {12}$
Conclusion:  if an odd prime $p$ divides $a^2+3b^2$ with coprime $a,b$ then $p\equiv 1,7\pmod {12}$.  Conversely, any such prime divides an expression of the desired form.
