Break It Down

Relevant wiki: Diophantine Equations - Solve by Factoring

$4^{a}-b^{2}=(2^{a}+b)(2^{a}-b)=c$ . Since $c$ is a prime, $2^{a}-b$ must be equal to $1$ , and so we get that $b=2^{a}-1$ .

We now get that $c=2^{a+1}-1$ by substituting $b$ into the factored version of the original equation.

This means that $b$ and $c$ are Mersenne Primes , more importantly, consecutive Mersenne Primes.

Either $a$ or $a+1$ , will be even, and in this case $2^{2j}-1=p$ for some integer $j$ and some prime $p$ . This can be factored once more into $(2^{j}+1)(2^{j}-1)=p$ . Once again, since $p$ is prime this means that $2^{j}-1=1$ , so $j=1$ and either $a,a+1=2$ .

If $a+1=2$ , then $a=1$ , which is not prime. So $a=2$ is the only possible solution for our initial equation. Solving for the other two variables yields $a=2,b=3,c=7$ , all of which are primes.

This is our only ordered triple, so our answer is $2+3+7=\boxed{12}$ .

Desenvolvendo,

$\\ \mathsf{4^a - b^2 = c} \\\\ \mathsf{(2^2)^a - b^2 = c} \\\\ \mathsf{2^{2a} - b^2 = c} \\\\ \mathsf{(2^a + b) \cdot (2^a - b) = c \cdot 1, \qquad \underline{c} \ \acute{e} \ primo.}$

Comparando os factores, $\begin{cases} \mathsf{2^a - b = 1 \qquad (i)} \\ \mathsf{2^a + b = c \qquad (ii)} \end{cases}$ .

Ora, da equação $\mathsf{(i)}$ , tiramos que $\mathsf{2^a}$ e $\mathsf{b}$ são dois números consecutivos, não necessariamente primos. Por conseguinte, de $\mathsf{(ii)}$ , tiramos que a soma desses dois consecutivos resulta num número primo; isto nos leva ao seguinte questionamento: a soma de dois números consecutivos é primo? Sim, todavia, precisamos de dois primos consecutivos e isso só ocorre com 2 e 3. Isto posto,

$\\ \mathsf{2^a + b = c} \\\\ \mathsf{2^2 + 3 = c} \\\\ \mathsf{4 + 3 = c} \\\\ \boxed{\mathsf{c = 7}}$

Por fim, $\mathsf{2 + 3 + 7 = \boxed{\mathsf{12}}}$ .