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.

Here is a similar looking result.

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

Proof. The statement follows readily from the identity $$(2a-b-1)\binom{b}{3}-(2b-a-1)\binom{a}{3}\ =\ (a+b-2)\binom{a-b+1}{3}.$$