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Let $S$ be a set of all positive rational numbers $x$ such that $2x^2 - 1$ is a square, excluding $x=1$.

I am interested in computing as many as possible solutions in $S$ to either the following equations:

(1) $\qquad p^2 - 1 = (q^2 - 1)\cdot r^2$

(2) ...removed...

(3) $\qquad p^2 + q^2 = 1 + r^2$

What would a reasonable computational approach for finding solutions?

For the equation (1), I know one solution: $(p,q,r)=(\tfrac{373}{23}, \tfrac{85}{41}, \tfrac{205}{23})$ -- would it help to find more solutions?

EDIT: Equation (2) was not exactly the one I'm interested in. So I removed it.

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1 Answer 1

Your question is equivalent to looking for rational points on certain hypersurfaces in affine 3-space.

The equation $2x^2 - 1 = y^2$ describes a plane conic with a rational point $(x,y) = (1,1)$. This means you can parameterize its solutions by passing a line through $(1,1)$ and finding the other point of intersection: $$ x = \frac{t^2 - 2t + 2}{t^2 - 2} \qquad y = \frac{-t^2 + 4t - 2}{t^2 - 2}. $$

Now you can substitute this expression for $x$ in for $p$, $q$, and $r$ (using a different variable for $t$ each time) to arrive at an equation for a hypersurface. For example, using the variables $a,b,c$, your equation (2) becomes $$ (a^2 - 2a + 2)(b^2 - 2b + 2)(c^2 - 2) = (c^2 - 2c + 2)(a^2 - 2)(b^2 - 2). $$

At this point, I think I've transformed your question, but I don't immediately see how to produce rational points on these hypersurfaces.

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Thanks! I've tried this approach but I also have no idea what to do with the resulting high-degree Diophantine equations. I would probably give up if I had just an equation like the one you derived. But I hope that the original formulation may provide additional insights into the structure of solutions and somehow simplify their search. –  Max Alekseyev Nov 20 '12 at 23:03
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The next thing I would do is to start exploring the geometry of these surfaces to get a better sense of what to expect. For example, is the projective closure smooth? A smooth hypersurface in $\mathbb{P}^3$ of degree at least 5 is of general type, so the Bombieri/Lang conjecture implies that all but finitely many of its rational points will lie on rational and elliptic curves on the surface. If that's the case, then finding the rational points is likely to be tricky. –  Xander Faber Nov 20 '12 at 23:49

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