Here is a problem I made one night for my friends who love to solve mathematical puzzles.

^{2}+ 3 =

^{x3}⁄

_{18}.

You might want to stop here and think for a while if you want to solve this yourself. There are spoilers ahead.

In this problem, I have hidden a Diophantine equation of the form
a^{n} + b^{n} = c^{n}. The first clue to this
hidden form might be the equation with its terms rearranged.

x^{3} = 18y^{2} + 54

The right hand side is 2(9y^{2} + 3^{3}). This is of
the form 2(3a^{2}b + b^{3}) which is equal to (a +
b)^{3} − (a − b)^{3}. Now, the hidden form
must be clear. So, let me write down the solution to this problem
neatly.

y^{2} + 3 = ^{x3}⁄_{18}

⇒ x^{3} = 18y^{2} + 54

⇒ x^{3} = (y + 3)^{3} − (y −
3)^{3}

⇒ 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} has got a trivial solution
a = b = 0 and no other solutions for positive integers a, b and c when n
> 2. Now, let us examine x^{3} + (y − 3)^{3} =
(y + 3)^{3} with the help of Fermat's Last Theorem. A trivial
solution in this case is ruled out because there is no value of y for
which y − 3 = y + 3 = 0.

However, if we consider y − 3 = 0, or y = 3, we get an equation that is no longer of the form required by Fermat's Last Theorem.

x^{3} = 6^{3}

⇒ x = 6

If we consider y + 3 = 0, or y = −3, again we get

x^{3} + (−6)^{3} = 0

⇒ x = 6.

Hence, the integral solutions to the equation are:

x = 6

y = ±3

## 1 comment

## MediumOne said:

Nice!

I would never thought of this: