Prove that $a+b+2\sqrt{ab+c^2}$ cannot be a prime number

Question

Prove that the number $$a+b+2\sqrt{ab+c^2}$$ cannot be a prime number for any positive integer numbers $a,b,c$.
My attempt:
Suppose that $p=a+b+2\sqrt{ab+c^2}$ is a prime. WLOG assume that $a \geq b$. From the equality we have $$\left(a+b\right)^2+p^2-2p\left(a+b\right)=4ab+4c^2 \Leftrightarrow \left(a-b-2c\right)\left(a-b+2c\right)=p\left(2a+2b-p\right)$$ Since $0<a-b+2c<a+b+2c<p$, we must have $p |\left(a-b-2c\right)$. But then I don't know how to continue the way. Please help me. Thank you.

Answer 1

Your work has almost finished the problem. $p|(a-b-2c)$ is the key property.

If $a-b-2c>0$, then $a-b-2c\ge p$. Hence $a\ge b+2c+p$, a contradiction since $a<p$.

If $a-b-2c<0$, then $a-b-2c\le -p$. Hence $a+p\le b+2c$, so $2a+b+2\sqrt{ab+c^2}\le b+2c$, so $a+\sqrt{ab+c^2}\le c$, again a contradiction.

Finally, if $a-b-2c=0$, then $a=b+2c$. But then $p=(b+2c)+b+2\sqrt{ab+c^2}$ is even.