TITLE: Are the only $b$ such that $b^2 + 4c$ and $b^2 - 4c$ are both perfect squares primes of the form $4k + 1$?
QUESTION [5 upvotes]: And of course, multiples of primes of the form $4k + 1$
$b, c, k$ are positive integers.
Ran into a math problem essentially involving this, just curious. I observed it to hold for the first $50$ integers, but not sure past that nor how to prove it.
Does it have to do with Fermat's "Christmas" theorem ($p = x^2 + y^2$ iff $p \equiv 1 \pmod 4$)?

REPLY [4 votes]: The number $b$ need not be prime. We have $65^2+3696=89^2$ and $65^2-3696=23^2$.
A boring kind of example is $10^2-96=4$, $10^2+96=14^2$.   
Remark: One can generate arbitrarily many non-boring  examples using $(r,s,b)$ such that $(r,s,b)$ is a primitive Pythagorean triple and $b$ is not prime. Any such example requires $b$ to have at least two not necessarily distinct prime divisors of the form $4k+1$.