Fall 2016, problem 31

Do there exist natural numbers $n,k$ with $1\le k\le n-2$ such that $$\binom{n}{k}^2+\binom{n}{k+1}^2=\binom{n}{k+2}^4 $$


5 years ago

This is equivalent to ${\dbinom{n}{k+2}}^{2}\frac{\left(k+2\right)^{2}}{\left(n-k-1\right)^{2}}+\frac{\left(k+1\right)^{2}}{\left(n-k\right)^{2}}\frac{\left(k+2\right)^{2}}{\left(n-k-1\right)^{2}} (1)$. However, $n-k\ge 2$ and $n-k-1\ge 1$ Thus, the $RHS$ of $(1)$ is $\le (k+2)^2+\frac{1}{4}(k+1)^2(k+2)^2\< \frac{1}{4}(k+2)^4$. This means that $\dbinom{n}{k+2}\< \frac{1}{2}(k+2)^2.$ This inequality holds for $n=k+2$ and $n=k+3$ but it clearly does not hold for $n\ge k+4$ Substituting $n=k+2$ into $(1)$ we see that there is no solution for $k$ that is a positive integer. The same is true for $n=k+3$.