Can 11 be represented by $a^2 - 3b^2$ where a and b are integers?

Question

I know the answer is no, just wan't to know how. From a similar question on the site I got that $a^2 - 3b^2$ should always equal a square modulo 3 which 11 is not. But I don't understand how to get to this conclusion. Please help

Answer 1

$$a^2-3b^2=11\implies a^2\equiv2\pmod3$$

But $a\equiv0,\pm1\pmod3\implies a^2\equiv0,1\pmod3$

Answer 2

There are two indefinite binary quadratic forms (well, classes) of discriminant $12.$ These are $x^2 - 3 y^2$ and $3 x^2 - y^2.$ The primes $2,3$ divide $12$ and are given separate consideration; both can be written as $3 x^2 - y^2.$

For any larger (positive) prime with Legendre symbol $(3|p) = 1,$ we have two choices.

If $p \equiv 3 \pmod 4,$ then $(p|3) = - (3|p) = -1,$ so $p \equiv 2 \pmod 3,$ and here $p \equiv 11 \pmod {12}.$ Such a prime is represented by $3 x^2 - y^2.$ So, $$11 = 12 -1, \; \; 23 = 27 - 4, \; \; \; 47 = 48 - 1, \; 59 = 75 - 16,$$ and so on.

If $p \equiv 1 \pmod 4,$ then $(p|3) = (3|p) = 1,$ so $p \equiv 1 \pmod 3,$ and here $p \equiv 1 \pmod {12}.$ Such a prime is represented by $ x^2 - 3y^2.$ So, $$13 = 16 -3, \; \; 37 = 49 - 12, \; \; \; 61 = 64 - 3, \; 73 = 100 - 27,$$ and so on.