Product of two number $a$ and $b$ How to prove that if $a, b$ are two positive integers so that $a^2+b^2=4100$ and $a<b<2a$ then $ab=2000$.

As pointed by another user, $a,b$ must be positive integers, otherwise it's easy to find a counterexample.
Let's find some limitation for $a$. From the condition we have:
$$4100 = a^2 + b^2 > 2a^2 \iff a^2 < 2050 \iff a < \sqrt{2050} \approx 45.27$$
Since $a$ is integer we can conclude $a\le 45$
Now for the lower bound:
$$4100 = a^2 + b^2 < a^2 + 4a^2 = 5a^2 \iff a^2 > 820 \iff a^2 > \sqrt{820} \approx 28.63$$
Now we found out that $29 \le a \le 45$
Now some modular arithemtic would come in use. Working modulo 8 we have:
$$a^2 + b^2 = 4100 \equiv 4 \pmod 8$$
Since $0,1,4$ are the only quadratic residues we can only make a sum of residues $4$ if $a^2 \equiv 2 \pmod 8$ and $b^2 \equiv 0 \pmod 8$ or other way around. 
For the first case let: $a=2k$ and $b=4n$, where $k$ is an odd integer. Now back to the limitation we have:
$$29 \le a = 2k \le 45 \iff 15 \le k \le 22$$
Now since $k$ is odd integer integer we have: $k \in \{15,17,19,21\}$
Checking all possibilities for $k$ wouldn't give us any solution.
Now check the case when $a=4n$ and $b=2k$, where $k$ is odd integer.
You'll get $a=40, b=50$ as the only integer solution satisfying the requirements. So we have: $ab = 40 \cdot 50 = 2000$.
Hence the proof.
