The diophantine equation X^2 - Y^2 - Z^2 = +- 1 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:54:19Z http://mathoverflow.net/feeds/question/114707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114707/the-diophantine-equation-x2-y2-z2-1 The diophantine equation X^2 - Y^2 - Z^2 = +- 1 Richard Bonne 2012-11-27T21:50:18Z 2012-11-28T14:04:19Z <p>Hi everybody. I'd like to know if the diophantine equation</p> <p>(1) $$X^2 - Y^2 - Z^2 = \pm 1$$</p> <p>has been studied and if the set of its solutions $(X,Y,Z)$ is known. I appreciate any reference. Thank you very much.</p> <p>P.S. If instead we look at the diophantine equation</p> <p>$$X^2 - Y^2 - Z^4 = \pm 1$$</p> <p>surely we can solve it imposing conditions on the solution of (1) so that Z be a square. However, is there a quicker method?</p> http://mathoverflow.net/questions/114707/the-diophantine-equation-x2-y2-z2-1/114712#114712 Answer by Will Jagy for The diophantine equation X^2 - Y^2 - Z^2 = +- 1 Will Jagy 2012-11-27T22:38:51Z 2012-11-27T22:49:24Z <p>It is also true that the automorphism group of the quadratic form is known. See <a href="http://mathoverflow.net/questions/110956/is-there-a-topograph-for-pythagorean-triples" rel="nofollow">http://mathoverflow.net/questions/110956/is-there-a-topograph-for-pythagorean-triples</a> and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements. </p> <p>I see, for your ordering you need to switch first and last elements to use these three matrices.</p> http://mathoverflow.net/questions/114707/the-diophantine-equation-x2-y2-z2-1/114720#114720 Answer by GH for The diophantine equation X^2 - Y^2 - Z^2 = +- 1 GH 2012-11-28T01:03:58Z 2012-11-28T14:04:19Z <p>In general, if $Q(x,y,z)$ is a nonsingular ternary quadratic form with integral coefficients, then the integral solutions of $Q(x,y,z)=n$ fall into finitely many orbits of the integral automorphism group of $Q$, and these orbits can be effectively determined. For your case there is a shortcut as follows.</p> <p>Write $a=x+y$, $c=x-y$, $b=z$, then your equation becomes $b^2-ac=\mp 1$. In the language of binary quadratic forms, this means that the form $au^2+2buv+cv^2$ has discriminant $\mp 4$. From classical theory it follows that after an invertible linear change of variables $u'=pu+qv$, $v'=ru+sv$ the form becomes $u'^2\pm v'^2$, i.e. there are $p,q,r,s\in\mathbb{Z}$ such that $ps-qr=1$ and $$au^2+2buv+cv^2=(pu+qv)^2\pm(ru+sv)^2.$$ Equating coefficients, $$a=p^2\pm r^2,\quad c=q^2\pm s^2,\quad b=pq\pm rs,$$ i.e. $$x=(p^2\pm r^2+q^2\pm s^2)/2,\quad y=(p^2\pm r^2-q^2\mp s^2)/2,\quad z=pq\pm rs.$$ To summarize, all integer solutions of $x^2-y^2-z^2=\pm 1$ are of this form, with integers $p,q,r,s$ satisfying $ps-qr=1$. Using congruences one can easily distinguish integer solutions from half-integer ones.</p> <p>As references, I suggest Rose: A course in number theory, and Cassels: Rational quadratic forms.</p>