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As the title stated, I am wondering the integers $x,y$ that satisfy the equation $x^2-y^3 = 23$.

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up vote 7 down vote accepted

This is a Mordell equation (or curve) and from this table (see E_+00023) no solution is known nor exists :

E_+00023: r = 0   t = 1   #III =  1
          E(Q) = {O}
          R =   1.0000000000
           0 integral points

(except the Tate-Shafarevic group #III the parameters correspond to the example 2 at the beginning of file i.e. there is no rational solution)

PROOF: Concerning the impossibility of integer solutions this was proved by Jonquières in 1878 and the (french) paper is available at Numdam. Starting from III of page 376 we find (adapted to your specific case $k=23$) :

We search solutions of $\quad y^3+23=x^2$.
Let's set $\ c:=3\ $ then $\ k=23=c^3-4\ $

Let's suppose that an even $y$ is solution of $\ y^3+3^3-4=x^2\ $ then the left part will be of type $8n-1$ while the right part will be of kind $8n+1$ (since $x$ must be odd).

We infer that $y$ must be odd and $x$ even so let's set $x:=2t$ and rewrite our equation as $$y^3+c^3=4(t^2+1)$$

This may be factorized as : $$(y+c)\underset{\text{odd}}{\underbrace{(y^2-cy+c^2)}}=4(t^2+1)$$

Since the second term must be odd ($o^2-o^2+o^2$) we deduce that the first one must be multiple of $4$. Since $\ c\equiv -1\bmod{4}\ $ we get $\ y=4n+1$.

The second factor $(y^2-3y+9)\equiv 1^2-(-1)1+1\equiv -1\bmod{4}\ $ i.e. it is of form $\ 4k-1$. But this factor should divide the $(t^2+1)$ factor at the right. I'll let you prove that this is not possible for a number of form $\ 4k-1$.

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