<s>[INST] Solve $ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $ with $a,b,c \in \mathbb{Z}$ I am trying to solve the Diophantine equation:
$$ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $$
Here's what it looks like if you expand, it's variant of the Pythagorean triples:
$$ a \times (a-1) + b \times (b-1) = c \times (c-1) $$
I was able to find solutions by computer search but this could have also been checked using the Hasse principle.
\begin{eqnarray*}
\binom{3}{2}+ \binom{3}{2}&=& \binom{4}{2} \\ \\
\binom{5}{2}+ \binom{10}{2}&=& \binom{11}{2} \\ \\
\binom{15}{2}+ \binom{15}{2}&=& \binom{21}{2} 
\end{eqnarray*}
and many others.  Is there a general formula for the $(a,b,c) \in \mathbb{Z}^3$ that satisfy this integer constraint.

>>> N = 25
>>> f = lambda a : a*(a-1)/2
>>> X = [(a,b,c,f(a) + f(b) - f(c)) for a in range(N) for b in range(N) for c in range(N)] 

>>> [(x[0],x[1],x[2]) for x in X if x[3] == 0 and x[0] > 1  and x[1] > 1 and x[2] > 1]

[( 3,  3,  4),  ( 4, 6, 7) , ( 5, 10, 11), ( 6,  4,  7), ( 6,  7,  9), 
 ( 6, 15, 16),  ( 7, 6, 9) , ( 7, 10, 12), ( 7, 21, 22), ( 9, 11, 14), 
 (10,  5, 11),  (10, 7, 12), (10, 14, 17), (10, 22, 24), (11,  9, 14), 
 (12, 15, 19), (12, 21, 24), (13, 18, 22), (14, 10, 17), (15,  6, 16), 
 (15, 12, 19), (15, 15, 21), (15, 19, 24), (18, 13, 22), (19, 15, 24), 
 (21,  7, 22), (21, 12, 24), (22, 10, 24)]

 [/INST] $$ (2a-1)^2 + ( 2b-1)^2 = 1 + (2c-1)^2  $$
Taking any $c,$
if the right hand side is not twice a prime, there will be nontrivial $a,b,$ while factoring the right hand side gives a method for finding them other than raw search. So that is one approach.
Parameterizing all uses the modular group  $SL_2 \mathbb Z$: take integers
$$ \alpha \delta - \beta \gamma = 1.  $$ We also demand $\alpha \beta + \gamma \delta $ odd. This means that one of $\alpha, \beta, \gamma, \delta$ is even and the other three odd. Then
$$ 2a-1 =  \alpha \delta + \beta \gamma $$
$$ 2b-1 = \alpha \beta - \gamma \delta $$
 $$ 2c-1 = \alpha \beta + \gamma \delta $$
Yet another approach is the automorphism group of the quadratic form. However, see page 124 in Magnus, Noneuclidean Tesselations and Their Groups, where Lemma 3.7 shows that this group also depends on the modular group. Well, actually 3.33 and 3.34 on the next page, now that I look more carefully.
I ought to emphasize that $x^2 + y^2 - z^2 = 0$ possesses a parametrization in just two variables, while $x^2 + y^2 - z^2 = 1$ jumps immediately to four variables with a condition, namely the modular group.  Worth adding $x^2 + y^2 + z^2 - w^2 = 0$ is naturally four variables, it is parameterized by the integral quaternions. 
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