Prime Number Sums

Question

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!

I know how to calculate the difference of two squares. $(a+b)(a-b)$. — R. G., Feb 03 '21 at 16:47
Clearly $2c-b < 2c+b$. Since $a$ is prime you have $2c+b=a$ and $2c-b=1$. — Crostul, Feb 03 '21 at 16:59
$c=3n$ $a=12n-1$ $b=6n+1$ $c=3n+1$ $a=12n+3$ $b=6n+1$ $c=3n+2$ $a=12n+7$ $b=6n+3$ — R. G., Feb 03 '21 at 18:06

score 1 · Answer 1 · answered Feb 03 '21 at 19:33

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: $a+b+c=12,19$

Prime Number Sums

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