Combination as Arithmetic Progression

If $p \ge 1$ and $1 \le q \le p-2$ , so that (in fact) $p \ge 3$ , and if the three binomial coefficients are in AP, then ${p \choose q} + {p \choose q+2} - 2{p \choose q+1} \; = \; 0$ and this expression is equal to $\begin{aligned} &\frac{p!}{(q+2)!(p-q)!}\left[ (q+1)(q+2) + (p-q)(p-q-1) - 2(q+2)(p-q)\right] \\ & = \frac{p!}{(q+2)!(p-q)!}\left[ 4q^2 + (8-4p)q + p^2 - 5p + 2 \right] \\ & = \frac{p!}{(q+2)!(p-q)!}\left[ (2q+2-p)^2 -p - 2\right] \end{aligned}$ Thus we require $(2q+2-p)^2 \; = \; p+2$ and hence $p+2$ must be a perfect square. Since $p \ge 3$ , the smallest possible value of $p$ is $\boxed{7}$ , and this works with $q=1$ (the relevant binomial coefficients are $7,21,35$ ).

A small extension of this argument shows that it is not possible to have ${p \choose q}\,,\,{p \choose q+1}\,,\,{p \choose q+2}\,,\, {p \choose q+3}$ all in AP. For this to happen, we would need $p+2 \; = \; (2q+2-p)^2 \; = \; (2(q+1) + 2 - p)^2$ and so $p+2 \,=\, u^2 = (u+2)^2$ where $u = 2q+2-p$ . The only way that two integers separated by $2$ can have the same square is when $u=-1$ , which would make $p+2=1$ , which is not possible.