Finding n and k that satisfies ${n \choose k} = 517 $

How do I find all the integers n and k that satisfies ${n \choose k} = 517 $ ? How do I approach this question?

The idea is to use the fact that for each $n$, the binomials $\binom{n}{0},\ldots,\binom{n}{\lfloor n/2 \rfloor}$ form an increasing sequence. We can assume that $k \leq n/2$. If $k \geq 2$ then $\binom{n}{k} \geq \binom{n}{2}$, and so, since $\binom{33}{2} = 528$, we must have $n \leq 32$. Checking $\binom{n}{k}$ for all $n \leq 32$, we find no solution. Hence we are left only with the trivial solutions $$ \binom{517}{1} = \binom{517}{516} = 517.$$ In fact, as Jorge Fernández correctly mentions, since the prime 47 is a factor of 517, we must have $n \geq 47$, and so no checking is needed.

• ${n \choose 0}$ is always 1, there is no solution
• ${n \choose 1} = n$, which gives the solution ${517 \choose 1}$
• ${n \choose 2} = \frac{n(n-1)}{2}$, which leads to the $n^2-n-1034=0$ quadratic equation, which doesn't have an integer solution.
• ${n \choose 3} = \frac{n(n-1)(n-2)}{6}$. But ${15 \choose 3}=455$, and ${16 \choose 3}=560$, so there is no solution again.
• Similarly, hogy ${n \choose 4}$, we got ${12 \choose 4}=495$ and ${13 \choose 4}=715$.
• For ${n \choose 5}$, ${11 \choose 5}=462$ and ${12 \choose 5}=792$. Here is no solution, too.
• For ${n \choose 6}$, ${11 \choose 6}=462$ (just as above), but ${12 \choose 6}=924$.
• From this point on, if we found a solution ${n \choose k}$, we had to also find it before. (Why?)

Thus the only solutions are ${517 \choose 1}$ and ${517 \choose 516}$.