The prime numbers a, b and c are such that a+b^2 =4 c^2 . Determine...

$$\begin{aligned} & \because 4 c^2-b^2=a \\ & \Rightarrow(2 c-b)(2 c+b)=a \\ & \Rightarrow 2 c-b=1 \text { and } 2 c+b=\mathrm{a} \\ & \Rightarrow 4 c=a+1 \text { and } 2 b=a-1 \end{aligned}$$ If $c \geq 5$ than $c$ must be of the type $6 \lambda \pm 1$ for $\lambda \in N$ Let $c=6 \lambda \pm 1$, then $a=24 \lambda+3$ or $24 \lambda-5 \because 24 \lambda+3$ is not a prime, so $c=6 \lambda-1$ and $a=24 \lambda-5$ then $b=\frac{24 \lambda-6}{2}=12 \lambda-3$, which is not a prime. So $c$ cannot be greater than 5 . If $c=3$ then $a=7$ and $b=5$ and If $c=2$ then $a=7$ and $b=3$ Sum of possible values of $a+b+c=31$