Find all pairs of integers $(x, y)$ such that $y^{2}=x^{3}+16$ (Factorization)
We have $(y-4)(y+4)=x^{3}$. Let $y-4=a$. If $2 \mid a$, then $2 \mid x$, so $8 \mid x^{3}=a(a+8)$, hence $a \equiv 0,4[8]$. If $a \equiv 4[8], 4 \mid a$, thus $4^{3} \mid x^{3}=a(a+8)$, so $a(a+8) \equiv 0[64]$, but $a=4+8 a^{\prime}$, thus $a(a+8)=64 a^{\prime 2}+64 a+16+32+64 a^{\prime} \equiv 48[64]$. This is absurd. Therefore, $8 \mid a$. Otherwise, $2 \nmid a$. Thus, $a \wedge(a+8)=a \wedge 8=8$ or 1. We have, if $a=8 b$: then there exists $y$ such that $x=4 y$, hence $b(b+1)=y^{3}$ and $b \wedge(b+1)=1$, so $b$ and $b+1$ are both cubes: necessarily $b=0$, thus $(x, y)=(0,4)$. If $a \wedge(a+8)=1$: then $a$ and $a+8$ are both cubes. Necessarily $a=0$, which is absurd. The only solution is therefore $(x, y)=(0,4)$.