Proof that $y^2=x^3+21$ has no integral solution with elementary methods? I tried to prove that $$y^2=x^3+21$$ has no integral solution in the way as shown here : http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf
My work so far : $x$ cannot be even because we would have $y^2\equiv 5\mod 8$ , which is a contradiction. Hence, $x$ is odd and $y$ is even. We have $$x\equiv x^3=y^2-21\mod 8$$, which gives the possibilities $x=3\ ,\ y=0\ or \ 4$ and $x=7\ ,\ y=2\ or\ 6$ modulo $8$.
Now I tried various factorizations to get a contradiction with quadratic residues :
$1)$ : $y^2+6=x^3+27=(x+3)(x^2-3x+9)$
$2)$ : $y^2-13=x^3+8=(x+2)(x^2-2x+4)$
$3)$ : $y^2-20=x^3+1=(x+1)(x^2-x+1)$
but in no case I found a contradiction with quadratic residues.

Can this particular equation be shown to be non-solveable in integers in the described way ?
How can I find systematically a suitable factorization in general leading to the desired contradiction, assuming the equation actually has no integral solution ?


This will not be a complete solution. Note that from your equations 1) and 3), two of the three possible cases of $x \pmod{3}$ can be eliminated:
If $x \equiv 0 \pmod{3}$, then from 1), $y^2 + 6 \equiv 0 \pmod{27} \longrightarrow y^2 \equiv 21 \pmod{27}$, which is impossible.
If $x \equiv 2 \pmod{3}$, then from 3), $y^2 - 20 \equiv 0 \pmod{9} \longrightarrow y^2 \equiv 2 \pmod{9}$, which is impossible.
Now for the apparently more difficult case of $x \equiv 1 \pmod{3}$. You have already shown that $x \equiv 3 \pmod{4}$, which means that $x \equiv 7 \pmod{12}$ is the only remaining possibility. Unfortunately, I can’t see how to carry this argument any further.
