$\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? I suspect there is some simple proof that no such p, q can exist, but I haven't been able to find one.


Solving the Pell equation gives candidates for p^2=x and q=y, with x=y=1 as the first candidate solution and subsequent ones given by x'=3x+4y, y'=2x+3y; chances of a prime square seem vanishingly unlikely as x increases, but I don't have a proof.


Meta: how do you search for a question like this? I looked for a searching HOWTO here and on meta, and couldn't find one. That the search appears to strip '^' and '=' makes it all the harder.



Are there primes p, q  such that p^4+1 = 2q^2 ?
This is not my solution, but I don't remember where I learned it.


Square both sides, subtract $4p^4$, and divide by 4 to obtain
$({p^4-1\over 2})^2=q^4-p^4$.


However, $z^2=x^4-y^4$ has no solutions in non-zero integers.
This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem.
The proof uses the representation of Pythagorean triples and infinite
descent. 


So you must have $p=\pm 1$. 


