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 $ab=2000$.
 A: 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.
A: Note that $4100=41\times 100$. As 41 is a prime number 1 mod 4, by Fermat's theorem can be expressed in only one way, viz: $41 = 4^2 +5^2$.
Now work with a smaller problem of expressing 100 as sum of two squares:
and then use the identity $(a^2+b^2)\times (c^2+d^2)\equiv (ac-bd)^2 + (ad+bc)^2 $. In this case it becomes $(4c-5d)^2 + (4d+ 5c)^2$ with $c^2+d^2=100$.
A: We can also view the conditions trigonometrically.  We are looking for integer coordinates $ \ (x,y) \ $ on the circle $ \ x^2 + y^2 = 4100 \ $ for which the line from the origin through the point(s) make(s) an angle to the positive $ \ x-$ axis such that $ \ \tan \theta = \ \frac{y}{x} \ $ is rational.
So if we require $ \ \tan \theta = \ \frac{p}{q} \ , $ we have $ \ \sin \theta = \ \frac{p}{\sqrt{p^2 + q^2}} \ , \ \cos \theta = \ \frac{q}{\sqrt{p^2 + q^2}} \ , $ and thus
$$ 2 \ \sin \theta \ \cos \theta \ = \ \sin 2 \theta \ = \ \frac{2 \cdot p \cdot q }{p^2 + q^2} \ = \ \frac{2 \cdot 2000 }{4100} \ = \ \frac{40}{41} \ . $$
Use of trigonometric identities, or even just a calculator, leads us to $ \ \tan \theta = \frac{4}{5} \ . $  The graph below is not intended to serve as "proof-by-picture", but simply to show what the conditions look like.

This serves to confirm the conclusions discussed by Stefan4024 (and the tangent value lurks in the last equation shown by P Vanchinathan).  It is also interesting that $ \ (31, 56) \ $ and $ \ (36, 53) \ $ come really  close (the sums-of-squares are 4097 and 4105, respectively), but only one point is a "perfect fit".
