# From Diophantus to Fermat

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

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