From Diophantus to Fermat

From Diophantus to Fermat

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

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 realize that this is a Diophantine equation of the form an+bn=cn. Here is how the equation looks after rearranging the terms:

x3=18y2+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(9y2+33). This is of the form 2(3a2b+b3) where a=y and b=3. Now 2(3a2b+b3)=(a+b)3−(a−b)3. Using these details, we get

x3=18y2+54⟺x3=2(9y2+33)⟺x3=(y+3)3−(y−3)3⟺x3+(y−3)3=(y+3)3.

From Fermat's Last Theorem, we know that an equation of the form an+bn=cn 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:

x>0,−3≤y≤3.

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 x3+(y−3)3=(y+3)3 would contradict Fermat's Last Theorem. Therefore y≤3 must hold good. Further since for every solution x and y, there is also a solution x and −y,−y≤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:

x=6y=±3.

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