How can we solve $y^2=x^3+23$ without trial and error? $$y^2=x^3+23$$
Are there any easy ways to solve this problem with number theory, abstract algebra, etc.?
(trial and error for mods by the way)

$$ y^2 + 4 = x^3 + 27 = (x+3)(x^2 - 3 x + 9)$$
If $x$ were even, we would have $y^2 \equiv 3 \pmod 4,$ impossible.
So, $x$ is odd, $y$ is even. Note that $y^2 + 4 $ is divisible by $4.$ Since $x^2 - 3 x + 9$ is odd, this means $x+3$ is divisible by $4.$ That is $x \equiv 1 \pmod 4.$
As a result, since $-3 \equiv 1 \pmod 4,$ we have $-3 x \equiv 1 \pmod 4,$ so
$$ x^2 - 3 x + 9 \equiv 3 \pmod 4.  $$
Finally, this means that there is some prime $q \equiv 3 \pmod 4$ such that
$q | (x^2 - 3 x + 9).$ 
The contradiction comes because that means $q |(y^2 + 4),$ this being impossible. Indeed, the lemma is that, given prime $q \equiv 3 \pmod 4,$ if $q |(u^2 + v^2),$ then both $q|u$ and $q |v.$ In this case, we have $v^2 = 4$ and $v = 2,$ and we cannot have $q | 2.$
$$ \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$ 
