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)
 A: $$ y^2 + 4 = x^3 + 27 = (x+3)(x^2 - 3 x + 9)$$
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 $q | 2.$
$$ \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$ 
A: If you consider the equation mod 4, we have $y^2-x^3 \equiv 3 \bmod 4$. Since $y^2 \equiv (0,1)$ and $x^3 \equiv (0,1,3)$, we must have $y$ even and $x\equiv 1 \bmod 4$.
Then setting $y = 2m$ and $x=4k+1$, we can rewrite:
$$ \begin{align}
(2m)^2 &= (4k+1)^3+23\\
4m^2 &= 64k^3+48k^2+12k+1+23\\
m^2 &= 16k^3+12k^2+3k+6\\
\end{align}$$
... and that doesn't obviously get us much further forward.
So I tried a little hunting around for other modular bases - $96$ is relatively sparse for solutions, and that gave me a chance to brute-force my way to a non-solution... $x=4489, y=300763$. (And managed to fool myself, even though I knew that $y$ needed to be even...)
Having that near-miss solution might provide some way forward for either finding real  solutions or closing the door...
