Binomial coefficients that sum to a perfect square

Question

I was wondering about a property of a sum I recently saw. $${8\choose2}+{9\choose 2}+{15\choose2} + {16\choose2}=17^2$$ And if we increment the terms $${9\choose3}+{10\choose 3}+{16\choose3} + {17\choose3}=38^2$$ Is it coincidence? I expanded them out into factorials but that did not reveal anything to me.

Answer 1

I began to explore this numerically in Mathematica. For $n\ge6$, let $$f(n) = \sqrt{\binom{n}{n-6}+\binom{n+1}{n-6}+\binom{n+7}{n-6}+\binom{n+8}{n-6}}$$

What I did was evaluated this expression for $n \in [6,10000]$. Any integral result was given a $1$ and all non-integers received a $0$. The total was summed to count the number of integral results...

f[n_, k_] := Sqrt[Binomial[n, k] + Binomial[n + 1, k] + Binomial[n + 7, k] + Binomial[n + 8, k]]

Total[Table[If[f[n, n - 6] \[Element] Integers, 1, 0], {n, 6, 10000}]]

The result was $3$. Which just gives the base case, as well as the two provided by the OP. Hope this helps as a starting point.