# Diophantine Equation Without Using Fermat's Last Theorem

I'm having trouble with a problem. The problem asks me to solve the equation $(x+1)^4-(x-1)^4=y^3$ in integers. I found out that the only integer solution is $(0,0)$. I found this answer by setting $x$ as $a^3$ and $x^2+1$ as $b^3$. After doing that, I got the equation $a^6+1^n=b^3$, which assures me that there is no other solution than $(0,0)$ by Fermat's Last Theorem. However, I just realized that I am not supposed to use Fermat's Last Theorem. So far, I have simplified the equation to $8x^3+8x=y^3$. Please help me prove that the only integer solution is $(0,0)$ without using Fermat's Last Theorem.

Consider $y^3=8x(x^2+1)$. Then $y$ must be even (say $y=2z)$. Then we get $z^3=x(x^2+1)$. Then we have $$z^3-x^3=x.$$ OR $$(z-x)(z^2+zx+x^2)=x$$ But the factor $|x^2+zx+z^2| \geq |x|$. So for this to be true $x=0=z$.

• Why is |x2+zx+z2|≥|x|? – Ajit Kadaveru Aug 5 '15 at 5:41
• @AjitKadaveru note that $z$ and $x$ are of same sign. This means $x^2+xz+z^2 \geq 0$. Moreover $x \in \mathbb{Z}$ thus $x^2 \geq x$. – Anurag A Aug 5 '15 at 15:53

Assume the contrary. Let $(x,y) \ne (0,0)$ be a non-zero solution to the equation

$$y^3 = 8x(x^2+1)\tag{*1}$$

Since RHS is even, so does LHS and $y$. Let $y = 2Y$, we have

$$Y^3 = x(x^2+1)$$

Since $\gcd(x,x^2+1) = \gcd(x,1) = 1$, both $x$ and $x^2 + 1$ are cubes.

Let $X, Z > 0$ be their cubic roots, we have

$$x = X^3,\; x^2 + 1 = Z^3 \quad\text{ and }\quad Y = XZ$$

This implies $$Z^3 = (X^2)^3 + 1 \implies Z^3 > (X^2)^3 \implies Z > X^2 \implies Z \ge X^2+1.$$ and hence $$1 = Z^2 - X^6 \ge (X^2+1)^3 - X^6 = 3X^4 + 3X^2 + 1\implies X = 0$$

This in turn implies $(x,y) = (0,0)$, contradict with our assumption.

As a result, the equation $(*1)$ has only $(0,0)$ as its solution.