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I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$

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I think writing $(x+iy)(x-iy) = (z-1)(z+1)$ might help. –  Joel Cohen Oct 22 '11 at 22:52
    
You are not trying to solve Project Euler problem 224, are you? projecteuler.net/problem=224 –  starblue Oct 23 '11 at 13:15
    
@starblue. No, it came up when I tried to make up an example in hyperbolic geometry. –  TCL Oct 23 '11 at 16:55
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2 Answers

Note: This is not a complete answer. First, I want to point out that both $x$ and $y$ must be even.

We can trivially get one infinitely family of solutions from the following: Consider triplets of the form $$(x,y,z)=(2r^2,2r,2r^2+1).$$ Then they satisfy the above as $$(2r^2 )^2 +(2r)^2+1=(2r^2+1)^2.$$ This gives rise to $$(2,2,3), \ (4,8,9), \ (6,18,19),\ \cdots$$

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The equation defines a conic with a rational point (0,0,1). The other rational solutions can be parametrized by lines through that point. As with $x^2+y^2 = z^2$ the integer solutions can be deduced from the rational ones, and the quadratic form $x^2+y^2 - z^2$ has a large linear symmetry group allowing one to move between solutions. In these respects the theory is the same as the one for Pythagorean triples.

There is a difference in the structure of the symmetry group. For $a=0$ the hyperboloid $x^2+y^2 - z^2 = a$ becomes a cone, with additional scaling symmetries in addition to the linear transformations of the hyperboloid; in effect the $O(2,1)$ symmetry group of the rational solutions collapses into a product of scalings and circle isometries. For integer solutions, with Pythagorean triples there is a reduction to the case of primitive triples, but when $a \neq 0$ there is a bound on the shared factors of $(x,y,z)$, and for $a = \pm 1$ all integer solutions are primitive. The organization of Pythagorean triples using several 3x3 integer matrices as transformations connecting different solutions does use the $O(2,1)$ structure, and the solution set for the equation with $a \neq 0$ could be presented in the same way (possibly with more than one connected component when $|a| > 1$).

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