# From Diophantus to Fermat

By Susam Pal on 12 Jan 2011

Here is a puzzle I created recently for my friends who love to indulge in recreational mathematics:

Find all integer solutions to the equation $y^2 + 3 = \frac{x^3}{18}.$

It does not take long to realize that this is a Diophantine equation of the form $$a^n + b^n = c^n.$$ Here is how the equation looks after rearranging the terms: $x^3 = 18y^2 + 54.$

The right hand side is positive, so any $$x$$ that satisfies this equation must also be positive, i.e., $$x > 0$$ must hold good for any solution $$x$$ and $$y.$$

Also, if some $$y$$ satisfies the equation, then $$-y$$ also satisfies the equation because the right hand side value remains the same for both $$y$$ and $$-y.$$

The right hand side is $$2(9y^2 + 3^3).$$ This is of the form $$2(3a^2b + b^3)$$ where $$a = y$$ and $$b = 3.$$ Now $$2(3a^2b + b^3) = (a + b)^3 - (a - b)^3.$$ Using these details, we get \begin{align*} x^3 = 18y^2 + 54 & \iff x^3 = 2(9y^2 + 3^3) \\ & \iff x^3 = (y + 3)^3 - (y - 3)^3 \\ & \iff x^3 + (y - 3)^3 = (y + 3)^3. \end{align*}

From Fermat's Last Theorem, we know that an equation of the form $$a^n + b^n = c^n$$ does not have any solution for positive integers $$a,$$ $$b,$$ $$c,$$ and positive integer $$n > 2.$$ Therefore we arrive at the following constraints for any $$x$$ and $$y$$ that satisfy the equation: \begin{align*} & x > 0, \\ -3 \le & y \le 3. \end{align*}

We established the first constraint earlier when we discussed that $$x$$ must be positive. The second constraint follows from Fermat's Last Theorem. If there were a solution $$x$$ and $$y$$ such that $$x > 0$$ and $$y > 3,$$ then $$x^3 + (y - 3)^3 = (y + 3)^3$$ would contradict Fermat's Last Theorem. Therefore $$y \le 3$$ must hold good. Further since for every solution $$x$$ and $$y,$$ there is also a solution $$x$$ and $$-y,$$ $$-y \le 3$$ must also hold good.

Since $$y$$ must be one of the seven integers between $$-3$$ and $$3,$$ inclusive, we can try solving for $$x$$ with each of these seven values of $$y.$$ When we do so, we find that there are only two values of $$y$$ for which we get integer solutions for $$x.$$ They are $$y = 3$$ and $$y = -3.$$ In both cases, we get $$x = 6.$$ Therefore, the solutions to the given equation are: \begin{align*} x & = 6 \\ y & = \pm 3. \end{align*}