Prime Number Sums The prime numbers $a,b$ and $c$ are such that $a+b^2=4c^2.$ Determine the sum of all possible values of $a+b+c.$
My Attempt
$a+b^2=4c^2$.
$a=4c^2-b^2$.
$a=(2c+b)(2c-b)$.
After this, I tried testing cases but I'm not totally sure how to account for everything.
Thanks!

Note that $c=2,3$ yield solutions: $a=7,b=3,c=2 \text{ and }a=11,b=5,c=3$
As has been noted in the comments, $2c-b=1$ is necessary for $a$ to be prime.
For prime numbers $c\ge 5$, $c$ must have the form $6k\pm 1$. Note that for $c=6k-1$ we have $2c-b=1=12k-2-b \Rightarrow b=12k-3$. Since $3\mid 12k-3$, $b$ cannot be prime and primes $c$ of the form $6k-1$ can never give rise to solutions.
Prime numbers $c=6k+1$ yield $2c-b=1=12k+2-b \Rightarrow b=12k+1$, which may or may not be prime. In cases where $b$ is in fact prime, $a=2c+b=12k+2+12k+1=24k+3$. Since $3\mid 24k+3$, $a$ cannot be prime and primes $c$ of the form $6k+1$ can never give rise to solutions.
So the only solutions are those given first, in which the asked for sums are:  $a+b+c=12,19$
