Are there primes p, q  such that p^4+1 = 2q^2 ? $\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? 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.
 A: I don't know if there is a simple proof, but I know one which is easy to do because it lets a computer do all the work (but the work is perhaps complicated): you simply ask a computer to solve Y^2=2X^4+2 in integers for you, like this (in MAGMA, but other packages will do it too):
> IntegralQuarticPoints([2,0,0,0,2]);
[
    [ 1, 2 ],
    [ -1, 2 ]
]

so the only solution with p,q integers is p,q=+-1 and that's it.
A: I think -- correct me if I am wrong -- that it is known that the equation $x^4+1=Dy^2$ with given squarefree $D$ has at most one solution in integers, primes or no primes. See for example J. H. E. Cohn., Math. Comp. 66 (1997), 1347-1351.  (http://www.ams.org/journals/mcom/1997-66-219/S0025-5718-97-00851-X/home.html) The article cites an original proof by Ljunggren in 1942, which I can't find online.
A: 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 $p=\pm 1$. 
