10
$\begingroup$

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)]
$\endgroup$
7
$\begingroup$

After some algebra you can rewrite the equation as $$ a(a-1)=(c-b)(c+b-1) $$

Note that $c-b$ and $c+b-1$ have opposite parity. Conversely, if $p$ and $q$ are integers with opposite parity, then we can find integers $b,c$ such that $p=c-b$, $q=c+b-1$: namely, $b=\frac{q+1-p}{2}$ and $c=\frac{q+1+p}{2}$.

That is to say, all solutions to the equation can be obtained in the following way:

  • Choose some number $a$. Factor $a(a-1)$ into two numbers $p,q$ of opposite parity. Since $a(a-1)$ is always even, this just means $p$ and $q$ can't both be even.
  • Take $b=\frac{q+1-p}{2}$, $c=\frac{q+1+p}{2}$.

For example, we could take $a=16$. Then $a(a-1)=240$, which factors into two numbers with opposite parity in four different ways:

\begin{align} 1&\cdot 240\\ 3 &\cdot 80\\ 5 &\cdot 48\\ 15 &\cdot 16 \end{align}

These give four essentially different solutions to the equation: \begin{align} \binom{16}{2}+\binom{120}{2}&=\binom{121}{2}\\ \binom{16}{2}+\binom{39}{2}&=\binom{42}{2}\\ \binom{16}{2}+\binom{22}{2}&=\binom{27}{2}\\ \binom{16}{2}+\binom{1}{2}&=\binom{16}{2} \end{align}

A few notes:

  • Swapping $p,q$, or replacing them by $-p,-q$, gives a solution which is essentially the same if you remember that $\binom{n}{2}=\binom{1-n}{2}$.
  • The symmetry between $a$ and $b$ means that this method will produce each solution twice. For example, the third solution above ($\binom{16}{2}+\binom{22}{2}=\binom{27}{2}$) could also have been found by taking $a=22$, $p=11$, $q=42$.
$\endgroup$
1
$\begingroup$

$$ (2a-1)^2 + ( 2b-1)^2 = 1 + (2c-1)^2 $$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.