Solution of $\large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ with $n\geq 3$

Question

I found this question in an old problem set. There's no hint or solution mentioned.

For $n \geq 3$, prove or disprove the existence of $(x,y,z) \in \mathbb N^3, \large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ (of course $x\geq n$, $y\geq n$,$z\geq n$

I really have no clue of the answer, and I can't disprove it either...

Answer 1

$${10\choose3}+{16\choose3}={17\choose3},\qquad{132\choose4}+{190\choose4}={200\choose4}$$ and see the discussion at https://mathoverflow.net/questions/27305/a-binomial-generalization-of-the-flt-bombieris-napkin-problem