Here is a problem I made one night for my friends who love to solve mathematical puzzles.
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 an + bn = cn. The first clue to this hidden form might be the equation with its terms rearranged.
x3 = 18y2 + 54
The right hand side is 2(9y2 + 33). This is of the form 2(3a2b + b3) 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.
y2 + 3 = x3⁄18
⇒ x3 = 18y2 + 54
⇒ 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 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 x3 + (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.
x3 = 63
⇒ x = 6
If we consider y + 3 = 0, or y = −3, again we get
x3 + (−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: