Italian Mathematical Olympiad 1994 - Problem 2

Problem:
Find all integer solutions of the equation $y^2=x^3+16$.

Solution:
We immediately see that $(0,-4)$ and $(0,4)$ are solutions of the given equation. We show that there are no other solutions. Rewriting the equation, we have $$(y-4)(y+4)=x^3.$$ Since the difference of the two factors on the left hand side is $8$, then $\gcd(y-4,y+4) \in \{1,2,4,8\}$ and we have $$y-4=2^n a, \qquad y+4=2^n b$$ with $n \in \{0,1,2,3\}, a,b \in \mathbb{Z}^*$, $\gcd(a,b)=1$. If $n=1,2$, from $2^n(b-a)=8$, we must have $a,b$ odd, but since $2^{2n}ab=x^3$, this is impossible by Unique-Prime-Factorization Theorem. If $n=3$, then $b-a=1$, and from $2^6a(a+1)=x^3$, we have that $a(a+1)$ is a perfect cube, and since $\gcd(a,a+1)=1$, $a$ and $a+1$ must be both perfect cubes, clearly impossible. At last $n=0$, i.e. $\gcd(y-4,y+4)=1$ and both $y-4$ and $y+4$ are perfect cubes whose difference is $8$. But this is impossible: in fact if there exist $k,m \in \mathbb{Z}$ such that $y-4=k^3$ and $y+4=m^3$, then $k^3$ and $m^3$ have the same parity and they are relatively primes, so they must be odd, and also $k$ and $m$ must be odd. Therefore, $$(m-k)(m^2+mk+k^2)=8$$ and by parity we force $$\begin{array}{ccc} m^2+mk+k^2 & = & 1 \\ m-k & = & 8, \end{array}$$ which implies $m^2+k^2=22 \equiv 6 \pmod{8}$, contradiction.