$2017$ is the first prime that satisfies the following three conditions:
$p$ can be written as $a^2+21b^2$ for integers $a,b$; in this case, $2017=41^2+21\cdot 4^2$.
$p$ can be written as $c^2+24d^2$ for integers $c,d$; in this case, $2017=29^2+24\cdot 7^2$.
$p$ satisfies the following modular congruence: $p \equiv 3 \mod 53$.
What is the second such prime?
The puzzle can be solved without the use of computers. There is one exception: if you suspect a number to be prime, then you can check that using a computer. However, in your final answer, you shouldn't use the prime number checker more than once.
A simple calculator is allowed, but not necessary to solve the problem; all calculations that have to be made, are relatively easy. If it is necessary to make a lot of calculations for your solution, try to find something clever that avoids these calculations. Good luck!
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