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Is there any literature / reference available for solving the Diophantine Equation $x^2 - y^2 - z^2 = d$?

I looked for it and found the following that were close, but not exactly the equation given here:

Are there any methods available to solve the equation $x^2 - y^2 - z^2 = d$?

vvg
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  • An obvious solution would be to set $z = 1$ and factor $d + 1$. How does one generate other solutions from this? – vvg Sep 24 '20 at 19:08
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    This is $(x+y)(x-y)=z^2+d$. For any $z$, this is soluble iff $z^2+d\not\equiv2\pmod4$, and solving it is essentially factoring $z^2+d$. – Angina Seng Sep 24 '20 at 19:42
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    The automorphism group or your quadratic form is large. The original reference is in German by Fricke and Klein, 1897. Anyway, you are asking about travelling from one solution to another. On this site, this is usually called Vieta Jumping. Here, the integer automorpisms are not so simple, but do have a nice decription. – Will Jagy Sep 24 '20 at 20:20
  • https://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2/789929#789929 – individ Sep 25 '20 at 04:22

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These are all the three by three matrices of determinant $+1$ that take an integer solution, $x,y,z$ written as a column vector, to a new one. On this site, I mostly see Vieta Jumping. This needs a little more work. The reference is pages 123-125 in Noneuclidean Tesselations and Their Groups by Wilhelm Magnus. It appeared first in Fricke and Klein (1897), especially pages 533-535. Fricke and Klein has been translated and that is available for sale from the AMS. The format is clever, all through the translation there are little marks on the side so you can tell what page in the German original is being presented. The subtitle of this book was

Erster Band: Die Gruppentheoretischen Grundlangen

First type, $ad-bc=1$ and $a + b + c + d \equiv 0 \pmod 2 \; , \;$

$$ \left( \begin{array}{ccc} \frac{1}{2} \left(a^2 + b^2 + c^2 + d^2 \right) & ab+cd &\frac{1}{2} \left(a^2 - b^2 + c^2 - d^2 \right) \\ ac+bd & ad+bc & ac-bd \\ \frac{1}{2} \left(a^2 + b^2 - c^2 - d^2 \right) & ab-cd & \frac{1}{2} \left(a^2 - b^2 - c^2 + d^2 \right) \\ \end{array} \right) $$

Second type $ad-bc=2$

$$ \left( \begin{array}{ccc} \frac{1}{4} \left(a^2 + b^2 + c^2 + d^2 \right) & \frac{1}{2}(ab+cd) &\frac{1}{4} \left(a^2 - b^2 + c^2 - d^2 \right) \\ \frac{1}{2}(ac+bd) & \frac{1}{2}(ad+bc) & \frac{1}{2}(ac-bd) \\ \frac{1}{4} \left(a^2 + b^2 - c^2 - d^2 \right) & \frac{1}{2}(ab-cd) & \frac{1}{4} \left(a^2 - b^2 - c^2 + d^2 \right) \\ \end{array} \right) $$

Will Jagy
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  • The parametric solution without need for factoring is what I was after. This solution addresses it. Great reference. – vvg Sep 25 '20 at 04:48