Are there primes p, q such that p^4+1 = 2q^2 ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:48:02Z http://mathoverflow.net/feeds/question/24609 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24609/are-there-primes-p-q-such-that-p41-2q2 Are there primes p, q such that p^4+1 = 2q^2 ? Hugo van der Sanden 2010-05-14T12:13:08Z 2010-05-14T15:26:18Z <p>$\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? I suspect there is some simple proof that no such p, q can exist, but I haven't been able to find one.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/24609/are-there-primes-p-q-such-that-p41-2q2/24612#24612 Answer by Kevin Buzzard for Are there primes p, q such that p^4+1 = 2q^2 ? Kevin Buzzard 2010-05-14T12:25:20Z 2010-05-14T12:25:20Z <p>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):</p> <pre><code>&gt; IntegralQuarticPoints([2,0,0,0,2]); [ [ 1, 2 ], [ -1, 2 ] ] </code></pre> <p>so the only solution with p,q integers is p,q=+-1 and that's it.</p> http://mathoverflow.net/questions/24609/are-there-primes-p-q-such-that-p41-2q2/24621#24621 Answer by Eben Freeman for Are there primes p, q such that p^4+1 = 2q^2 ? Eben Freeman 2010-05-14T14:01:44Z 2010-05-14T14:01:44Z <p>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. (<a href="http://www.ams.org/journals/mcom/1997-66-219/S0025-5718-97-00851-X/home.html" rel="nofollow">http://www.ams.org/journals/mcom/1997-66-219/S0025-5718-97-00851-X/home.html</a>) The article cites an original proof by Ljunggren in 1942, which I can't find online.</p> http://mathoverflow.net/questions/24609/are-there-primes-p-q-such-that-p41-2q2/24636#24636 Answer by Byron Schmuland for Are there primes p, q such that p^4+1 = 2q^2 ? Byron Schmuland 2010-05-14T15:26:18Z 2010-05-14T15:26:18Z <p>This is not my solution, but I don't remember where I learned it.</p> <p>Square both sides, subtract $4p^4$, and divide by 4 to obtain $({p^4-1\over 2})^2=q^4-p^4$.</p> <p>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. </p> <p>So you must have $p=\pm 1$. </p>