Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$ 
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$

I guess, that x cannot be even. because we can apply mod 4 test
say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$ 
but, this is not possible, since the square of an integer is congruent to $0$ or $1$ $\mod (4)$
so $x$ is of the form $4k+3$ or $4k+1$
$x=4k+3$ is also not possible for the same reason but for the last case the test fails.
any hints ?

There are no integer solutions.
You noted already that $x$ cannot be even, because the equation would then
fail mod $4$; so we need only consider $x$ odd.  Then $x^3+23$ is even,
so must be a multiple of $4$, whence $x \equiv 1 \bmod 4$.
We cannot obtain a contradiction from further congruence considerations.
But since $23 = 3^3 - 2^2$ we try adding $4$ on both sides, getting
$$
y^2 + 4 = x^3 + 27 = (x+3) (x^2-3x+9),
$$
and observe that $x \equiv 1 \bmod 4$ implies
$x^2 - 3x + 9 \equiv 1 - 3 + 9 \equiv  3 \bmod 4$.
Since also $x^2-3x+9 > 0$ we deduce that $x^2-3x+9$ has a prime factor
$p \equiv 3 \bmod 4$.  But this is impossible by quadratic reciprocity:
$y$ would be a square root of $-4 \bmod p$.  QED
(In fact mwrank reports that there aren't even any rational solutions,
but that's not an elementary proof.)
[Added later: See Keith Conrad's
Examples
of Mordell's Equation
for further examples of elementary but nontrivial proofs that certain
equations of this form $y^2 = x^3 + k$ have no integer solutions,
and also some examples where a nonempty list of solutions of $y^2 = x^3 + k$
can be proved complete.]
