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

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

If you want to think about this puzzle, this is a good time to pause and think about it. There are spoilers ahead.

It does not take long to realise 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. \( x \gt 0 \) 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 \[ x^3 = 18y^2 + 54 = 2(9y^2 + 3^3) = (y + 3)^3 - (y - 3)^3. \] Rearranging the terms, we get \[ x^3 + (y - 3)^3 = (y + 3)^3. \] 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 \( n \gt 2 \). We know that \( x \gt 0 \). Therefore \( y \gt 3 \) contradicts Fermat's Last Theorem. Thus the inequality \( y \le 3 \) must hold good. Further since for every solution \( x \) and \( y \), there is also a solution \( x \) and \( -y \), the inequality \( -y \le 3 \) must also hold good. Therefore values of \( x \) and \( y \) that satisfy the above equation must satisfy the following inequalities: \begin{align*} x \gt 0, \\ -3 \le y \le 3. \end{align*} 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: \[ x = 6, \qquad y = \pm 3. \]