For any integer \(n > 2\), the equation \(a^n + b^n = c^n\) has no positive integer solutions for \(a, b, c\).
Proof Outline
Case 1: When \(n=2\)
For the integer \(n=2\), there are infinitely many positive integer solutions. These are the well-known Pythagorean equation.
For example:
\[ 3^2 + 4^2 = 9 + 16 = 25 \] \[ 3^2 + 4^2 = 5^2 \]Another example with \(a=5\) and \(b=12\) is:
\[ 5^2 + 12^2 = 25 + 144 = 169 \] \[ 5^2 + 12^2 = 13^2 \]In both cases, we see that solutions exist when the power \(n=2\).
Case 2: When \(n > 2\)
For any integer \(n > 2\), the equation \(a^n + b^n = c^n\) has no positive integer solutions.
For example, if we consider \(n=3\), we would need to find positive integers \(a, b, c\) such that:
\[ a^3 + b^3 = c^3 \]Such non-zero integer numbers do not exist.
Similarly, for \(n = 4, 5, 6, \dots\), there are also no positive integer solutions.
Proved
