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$\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? 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.

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    $\begingroup$ If you're looking for a simple proof, you probably have to a) factor the left hand side over Z[i]; b) subtract 1 and factor the right hand side over Z[sqrt{2}], or c) multiply by 2 and use the classification of Pythagorean triples (or subtract a square and factor the right hand side, which is a difference of squares). $\endgroup$
    – Franz Lemmermeyer
    Commented May 14, 2010 at 12:31
  • $\begingroup$ Yes, $\mathbb{P}$ is the primes. That is standard isn't it? $\endgroup$
    – Hugo van der Sanden
    Commented May 14, 2010 at 12:32
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    $\begingroup$ @Franz: This is a variation of Pell's equation, so $p^2+q\sqrt2=(1+\sqrt2)^{2k+1}$ (it can be shown by modular argument that $k$ is multiple of 4). Thus (a) can't work, (b) is used in derivation of the above formula (there are useless recursions mentioned by the author), and for (c) I have no idea of what do you mean (how to relate the equation with Pythagorean triples?) $\endgroup$
    – Wadim Zudilin
    Commented May 14, 2010 at 13:45
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    $\begingroup$ Interestingly enough, the similar equation p^2 + 1 = 2q^4 DOES have an interesting solution, namely (p,q) = (239,13). This is related to Machin's identity pi/4 = 4 arctan(1/5) - arctan(1/239). (See Ribenboim's book on Catalan Conjecture for this.) See p.7-8 of www.math.wisc.edu/~ellenber/MCAV.pdf for an "explanation" of this solution via a congruence mod 5 between a weight-2 cuspform in conductor 1024 and an Eisenstein series. $\endgroup$
    – JSE
    Commented May 14, 2010 at 20:45
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    $\begingroup$ @Wadim: I can't see why a) cannot work. As for c), 2p^4 + 2 = (p^2+1)^2 + (p^2-1)^2 = (2q)^2. $\endgroup$
    – Franz Lemmermeyer
    Commented May 16, 2010 at 16:10

3 Answers 3

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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 $p=\pm 1$.

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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.

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    $\begingroup$ How can Magma be used with confidence, their code is not open. $\endgroup$
    – teil
    Commented May 14, 2010 at 13:39
  • $\begingroup$ Negative's point is a valid one, but I think Kevin's answer points to the fact that there is a standard algorithm to solve these kinds of problems. Maybe it is also implemented in other packages. $\endgroup$
    – Felipe Voloch
    Commented May 14, 2010 at 14:59
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    $\begingroup$ This is only semi-true. The outer code of SIntegralPoints in Magma can be viewed easily. As Felipe Voloch says in part, this lists the algorithm. Low-level functions (eg. Height) cannot be seen, though "open" is technically true as you can decompile. From the philosophy of science, no system is used with confidence, and only independent verification can be partially satisfactory. We still have at least two chip manufacturers, so hardware independence is possible. For using software, too many packages throw eggs into a large basket (like GMP), and so I see a scarcity of true independence often. $\endgroup$
    – Junkie
    Commented May 15, 2010 at 6:19
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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.

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  • $\begingroup$ Ah, excellent. That suggests x=y=1 is the only solution; I'll take a look at the paper to see if it constitues a proof. $\endgroup$
    – Hugo van der Sanden
    Commented May 14, 2010 at 14:07
  • $\begingroup$ Oh, it is the Ljunggren paper I'd need. $\endgroup$
    – Hugo van der Sanden
    Commented May 14, 2010 at 15:40

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