If $a,b,c$ are nonzero natural numbers that verify the relation $a^2+b^2=c^2$, show that $a^n+b^n<c^n$.

Question

PROBLEM

If $a,b,c$ are nonzero natural numbers that verify the relation $a^2+b^2=c^2$, show that $a^n+b^n<c^n$, where $n$ is a strictly natural number greater than $2$.

WHAT I THOUGHT OF

$a^2+b^2=c^2$ is the reminds me of the The Pythagorean Theorem.

We can say that $a,b,c$ can be the sides of a right angled triangle, with c being the hypotenuse.

Using the triangle inequality we can write that

$a+b>c$

$b+c>a$

$a+c>b$

I also thought that we can write $a^2+b^2=c^2$ as $a^2=c^2-b^2=(c+b)(c-b)$.

I don't know what to do forward. Hope one of you can help me.

Answer 1

Define $\alpha := a/c$ and $\beta := b/c,$ so that $\alpha^2 + \beta^2 = 1.$ This in particular means that both $\alpha$ and $\beta$ lie strictly between $0$ and $1,$ which implies that $$\alpha^n+\beta^n < \alpha^2+\beta^2 = 1 \quad \text{for all } n \geq 3.$$ Now multiply by $c^n$ to get the desired inequality.

Answer 2

In general, for Fermat's last theorem, for all triple $(a,b,c)$ with $a\lt b\lt c$ such that $a^2+b^2\gt c^2$ there exists an irrational $\theta$ such that $a^{\theta}+b^{\theta}=c^{\theta}$ and if $m=\lfloor(\theta)\rfloor$ then $$ 0\lt a+b-c\lt a^2+b^2-c^2\lt \cdots\lt a^m+b^m-c^m$$ and, now decreasingly, for all integer $k$ $$a^{m+k}+b^{m+k}-c^{m+k}\lt 0$$ You have besides, if $f(x)=a^x+b^x-c^x$, the formula $$f(x_0+3)=(a+b+c)f(x_0+2)-(ab+ac+bc)f(x_0+1)+abcf(x_0)$$ where $x_0$ is an arbitrary real.This formula can be generalized for $f(x_0+3h)$.

Example.-For $f(x)=17^x+19^x-20^x$ one has $m=\lfloor(\theta)\rfloor=7$ so $$\begin{cases}17+19-20\lt 17^2+19^2-20^2\lt\cdots \lt17^7+19^7-20^7\\ 17^{\theta}+19^{\theta}=20^{\theta}\\ 0\gt17^8+19^8-20^8\gt17^{8+k}+19^{8+k}-20^{8+k}\end{cases}$$ decreasing fastly for all integer $k$.