Find all natural number solutions to: $20x^2 + 11y^2 = 2011$

Question

I believe that the equation $$20x^2 + 11y^2 = 2011$$ describes an ellipse. I don't know how to solve for the $x,y \in \mathbb{N}$ that satisfy this equation.

Answer 1

Since we have $$2011-20x^2=11y^2\ge 0\Rightarrow 2011-20x^2\ge 0\Rightarrow x^2\le\frac{2011}{20}=100.55,$$ we have $$x^2=1,4,9,16,25,36,49,64,81,100.$$ Then, trying each of these to find a natural number $y$ such that $$y^2=\frac{2011-20x^2}{11}$$ gives us that there is only one solution $(x,y)=(10,1).$

Answer 2

The main thing is that 2011 is prime, easy enough to check. That means there is only one way to write it as $m x^2 + n y^2$ with positive integers. Since $2000 =20 \cdot 100 = 20 \cdot 10^2, $ we see that we can write it as $2011 = 20 \cdot 10^2 + 11 \cdot 1^2.$ As Brillhart mentions below, the part about primes is also proved on page 222 of Number Theory: An approach through history from Hammurapi to Legendre by Andre Weil.

The fact that there can be only one such way to write a prime number is due originally to Euler, but not written down well by his students after he became blind. A proof was written out by Lucas much later. This is from an article I like by John Brillhart: