#### Problem

The radical of $n$, $\text{rad}(n)$, is the product of distinct prime factors of $n$. For example, $504 = 2^3 \times 3^2 \times 7$, so $\text{rad}(504) = 2 \times 3 \times 7 = 42$.

Given any triplet of relatively prime positive integers $(a, b, c)$ for which $a + b = c$ and with $a \lt b \lt c$, it is conjectured, but not yet proved, that the largest element of the triplet, $c \lt \text{rad}(abc)^2$.

Assuming that this conjecture is true, prove that $x^n + y^n = z^n$ has no integer solutions for $n \ge 6$.

#### Solution

Let $a = x^n$, $b = y^n$, and $c = z^n$.

If $x$, $y$, and $z$ were coprime then the maximum radical of $abc$ would be $xyz$.

Therefore $\text{rad}(abc) \le xyz$, but as $z$ is the greatest in value, it follows that $\text{rad}(abc) \lt z^3$.

By the conjecture, $c \lt \text{rad}(abc)^2 \lt (z^3)^2$; that is, $c = z^n \lt z^6$.

Hence $n \lt 6$, and we conclude that $x^n + y^n = z^n$ has no integer solutions for $n \ge 6$.

The cases of $n$ = 3, 4, and 5 all have elementary proofs, so if this conjecture were true, it would provide for an elegant completion of the proof of FLT (Fermat's Last Theorem). Of course the importance of this conjecture not yet being proved cannot be overstressed. It may turn out that the proof of this conjecture is more difficult than the current proof for FLT. Also note that the proof of FLT does not provide proof of this conjecture. However, mathematicians are still encouraged to find a proof for this conjecture, and other results relating to the radical function, as its usefulness is far reaching into many other areas of current mathematical research.

Problem ID: 284 (23 Jul 2006)     Difficulty: 3 Star

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