Solving for $x=?$

Let $f: \R\to\R$ be the function defined by

$f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x$ .

To solve the equation $3^x +4^x = 5^x$ is equivalent to solving $f(x) = 1$ . It is clear that $x=2$ is a solution; to show that it is unique, we note that $f(x)$ is a decreasing function, since

$f'(x) = \ln(\frac{3}{5})(\frac{3}{5})^x + \ln(\frac{4}{5}) (\frac{4}{5})^x < 0\,$ for all $x\in\R$ .

Therefore, $f(x)\neq 1$ if $x\neq 2$ . $\square$

$3^2+ 4^2= 5^2$ is the only solution. We will show by induction that for $x ≥ 3$ we always have $3^x+4^x \le 5^x$

Let $x=3 \implies 3^3+ 4^3 \le 5^3$ ,

suppose it is true for $x=k,\implies 3^k+ 4^k \le 5^k,( k > 0)$

Now, we need to prove it is true for $x= k+1, \implies 3^{k+1}+4^{k+1} \le 3^k*3+ 4^k*4 \le 5^k *5 = 5^{k+1}.$

Now, similarly for $x \le 1 \rightarrow 3^x+ 4^x \ge 5^x \implies$ the only solution is when $x=2$ , we get $3^2+4^2=5^2$ .

Fermat's last theorem :)