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If asked to express the generic pythagorean triplets satisfying $a^2+b^2=c^2$ you would answer $$ a = k(r^2 - s^2), b = 2krs, c = k(r^2+s^2) $$ with $k,r,s \in \Bbb N$ and $r>s$ and $r \ne s \mod 2$ and $\gcd(r,s) = 1$. This generates every pythagorean triplet exactly once as the pair $(r,s)$ takes on its allowed values.

I'm looking to derive an analogous formula to generate every "penthagorean" triplet, that is, every instance where the sum of two pentagonal numbers is a pentagonal number. That is, find all $(a,b,c)$ satisfying $$ \frac{3a^2-a}2+\frac{3b^2-b}2=\frac{3c^2-c}2 $$

Deriving the formula for pythagorean numbers, using only elementary number theory such as considerations in mods $2$ and $4$ is so straightforward that I thought the analogous formula would be easy to obtain, but I am getting stuck.

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from pages 124,125 of Magnus. If we have integers $a,b,c,d$ with $ad-bc=1$ and $a+b+c+d \equiv 0 \pmod 2,$ and we take $$ \left( \begin{array}{ccc} \frac{1}{2} \left( a^2 + b^2 + c^2 + d^2 \right)&ab+cd&\frac{1}{2} \left( a^2 - b^2 + c^2 - d^2 \right) \\ ac+bd&ad+bc&ac-bd \\ \frac{1}{2} \left( a^2 + b^2 - c^2 - d^2 \right)&ab-cd&\frac{1}{2} \left( a^2 - b^2 - c^2 + d^2 \right) \\ \end{array} \right) \left( \begin{array}{c} x \\ y\\ z \\ \end{array} \right) = \left( \begin{array}{c} u \\ v \\ w \\ \end{array} \right) $$

THEN $$ u^2 - v^2 - w^2 = x^2 - y^2 - z^2 $$

At a minimum, the modular group takes a solution to your problem to another. Oh, multiply by 12 and complete the square, $$ (6p-1)^2 + (6q-1)^2 = 1 + (6r-1)^2 $$ so that $6r-1$ becomes either $x$ or $u$

The observations in Magnus (1974) go back to Fricke and Klein (1897)

I cannot recall whether this gives all solutions to $x^2 - y^2 - z^2 = -1.$ Worth experimenting. The 3 by 3 matrix has determinant 1 and should preserve $\gcd(x,y,z).$

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  • $\begingroup$ The relationship to the "missed it by that much" triples where $x^2+y^2 = z^2+1$ had escaped me. But I had tried last year to get the generic solution to that one, and could not do that either. $\endgroup$ – Mark Fischler May 25 '18 at 23:15
  • $\begingroup$ @MarkFischler depends what you want to do with it. You can switch to $rs-pq=2,$ take $x+1 = rs, x-1=pq,$ $z-y=pr, z+y = q s$ $\endgroup$ – Will Jagy May 26 '18 at 0:20
  • $\begingroup$ This provided a clue which allows me now to express a 3-parameter formula that I think might generate all triples of $x^2+y^2=z^2+1$. The formula is too long for a comment; I will add an answer instead. $\endgroup$ – Mark Fischler Jun 5 '18 at 21:12
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Here are a few one-dimensional parametric forms, if it helps. This was found by assuming the solution space was a line in $\mathbb{N}^3$.

$$ \begin{align} &(16x+4, 30x+7, 34x+8) \\ &(33x+4, 56x+7,65x+8) \\ &(7x+7, 24x+23, 25x+24) \\ \end{align} $$

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A $3$-parameter formula which always generates triplets of the form $$ x^2+y^2=z^2+1 $$ is (with $m,n,p \in \Bbb{Z}$)

$$ x = 2(mn+1)(mp-1)+1 \\ y = pm^2(n^2-1)+m(2np-n^2+1)+p-n \\ z= pm^2(n^2+1)+m(2np-n^2-1)+p-n $$

I'm not certain that this formula generates all $(x,y,z)$ triples satisfying $x^2+y^2=z^2+1$ or even that it generates all such triples that can be formed using the Fricke/Klein technique, but numerical exploration indicates that it seems to cover them all. Unlike the 3-parameter pythagorean formula, this formula does repeat triples, some in trivial ways but others by non-trivial coincidence.

Later observation - This 3-parameter formula does not generate all the triples satisfying $x^2+y^2=z^2+1$.

As noted by @Will Jagy, when all three elements $(x,y,z)$ of the triplet are of the form $6k-1$ you can form a penthagorean triplet $(\frac{r(3r-1)}2, \frac{s(3s-1)}2, \frac{t(3t-1)}2)$ using $$ (r = \frac{x+1}6, s =\frac{y+1}6, t = \frac{z+1}6) $$

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