Interesting note: The above problem is taken of this year's fourth round which is the last, national round. The third round's first problem was the following: Find all positive integers $a,b$ that solve \[ \binom{ab+1}{2} =2ab(a+b). \]German problem proposers often use variations/continuations of problems as of lately.

These should probably go in the High School Olympiads forum.

Nah, I don't really think so as that problem isn't all too hard.

$ \binom{2n}{n}=\sum_{0 \le i \le n} \binom{n}{i}^2$ thus$\sum_{0 \le i \le n} \binom{n}{i}^2=\frac{m(m^2-1)}{2} \le \frac{2n(4n^2-1)}{2}=n(4n^2-1)$ thus after dividing everything with $n$ we have $\sum_{0 \le i \le n}n \cdot \binom{n-1}{i}^2 \le 4n^2-1$ so again after dividing everything with $n$ we have $\sum_{0 \le i \le n} \binom{n-1}{i}^2 \le 4n-\frac{1}{n} <4n$ now it's just boring case work i.e bounding $n$ and by that we easily find all solutions I must say that I'm bit surprised that German MO isn't any harder (at least this problem, though it's nice)

^Well, first problems of the German MO are often very easy. There are some tough ones but in general, I guess German MOs are one the easier ones out there. This year's problems seemed to be especially easy, because in 11th grade, a gold medal wasn't even awarded for 36/40 points.