Here is a similar looking result.

**Lemma.** For any integers $a$ and $b$,
$$\gcd\left(\binom{a}{3}, \binom{b}{3}\right)\qquad\text{divides}\qquad (a+b-2)\binom{a-b+1}{3}.$$

**Proof.** For any integers $a$ and $b$, we have
$$(2a-b-1)\binom{b}{3}-(2b-a-1)\binom{a}{3}\ =\ (a+b-2)\binom{a-b+1}{3}.$$
The divisibility relation follows readily.