General integer solution for x^2+y^2-z^2=+/-1 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:26:04Z http://mathoverflow.net/feeds/question/65957 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65957/general-integer-solution-for-x2y2-z2-1 General integer solution for x^2+y^2-z^2=+/-1 Victor Kuliamin 2011-05-25T12:28:54Z 2011-06-15T15:52:28Z <p>How to find general solution (in terms of parameters) for diofantine equations x^2+y^2-z^2=1 and x^2+y^2-z^2=-1?</p> <p>It's easy to find such solutions for x^2+y^2-z^2=0 or x^2+y^2-z^2-w^2=0 or x^2+y^2+z^2-w^2=0, but for these ones I cannot find anything relevant.</p> <p>Victor Kuliamin</p> http://mathoverflow.net/questions/65957/general-integer-solution-for-x2y2-z2-1/65961#65961 Answer by Gerry Myerson for General integer solution for x^2+y^2-z^2=+/-1 Gerry Myerson 2011-05-25T12:50:37Z 2011-05-26T02:33:43Z <p>I believe the general solution to $x^2+y^2-z^2=1$ is $x=(rs+tu)/2$, $y=(rt-su)/2$, $z=(rs-tu)/2$, where $rt+su=2$. </p> <p>EDIT: Solutions to $x^2+y^2+1=z^2$ can be obtained by choosing $a$, $b$, $c$, $d$ such that $ad-bc=1$ and then letting $x=(a^2+b^2-c^2-d^2)/2$, $y=ac+bd$, $z=(a^2+b^2+c^2+d^2)/2$, though I'm not sure you get all the integer solutions this way. </p> <p>The rational solutions are a bit easier. $(0,0,1)$ is a (rational) point on the surface. The line $(0,0,1)+t(a,b,c)$ through that point intersects the surface again at $x=2ac/(a^2+b^2-c^2)$, $y=2bc/(a^2+b^2-c^2)$, $z=(a^2+b^2+c^2)/(a^2+b^2-c^2)$, giving all the rational points on the surface. </p> http://mathoverflow.net/questions/65957/general-integer-solution-for-x2y2-z2-1/66018#66018 Answer by Aaron Meyerowitz for General integer solution for x^2+y^2-z^2=+/-1 Aaron Meyerowitz 2011-05-26T03:47:49Z 2011-06-15T12:48:11Z <p>I think that the solutions to $x^2+y^2-z^2=-1$ are $x=RT-SU,y=RU+ST$ where $R^2+S^2-T^2-U^2=2$ then $z=R^2+S^2-1=T^2+U^2+1$ On the surface this looks similar to the solutions to the $+1$ case. However these are quite a bit rarer and depend on the locations of the primes. </p> <p>As we know, an integer can be uniquely written as $n=ab^2$ where $a$ (the squarefree part of $n$) is a product of distinct primes. $n$ can be written as a sum of two squares $n=j^2+k^2$ precisely when $a$ has no prime divisors of the form $4m+3$ (and we know in how many ways this can be done as well.) So the solutions depend on when we have 2 consecutive even numbers of this form. </p> <p>For example $292=73\cdot4^2$ and $290=2\cdot5\cdot329$ thus we know that there are expressions as a sum of two squares: $$292=6^2+16^2$$ $$290=1^2+17^2=11^2+13^2.$$ Running through the various possiblities gives these solutions for $R,S,T,U,x,y$ with $x^2+y^2-291^2=-1:$</p> <ul> <li>6, 16, 17, 1, 86, 278 </li> <li>16, 6, 11, 13, 98, 274 </li> <li>16, 6, 17, 1, 266, 118 </li> <li>16, 6, 13, 11, 142, 254 </li> </ul> <p>Certain families of solutions can be given. One is $x,y,z=2p,2p^2,2p^2+1.$</p> http://mathoverflow.net/questions/65957/general-integer-solution-for-x2y2-z2-1/67874#67874 Answer by Richard Borcherds for General integer solution for x^2+y^2-z^2=+/-1 Richard Borcherds 2011-06-15T15:52:28Z 2011-06-15T15:52:28Z <p>The question is the same as looking for points of norm 1 or -1 in the unimodular Lorentzian lattice $Z^{1,2}$. This has an infinite group of automorphisms, with an index 2 subgroup that is a Coxeter group generated by 3 reflections. This group acts transitively on the vectors of norm 1 and -1 if I remember correctly, so all solutions can be obtained from 1 particular solution by acting with this group. </p>