Proof that $(a^m+b^m)^{n}= (a^n+b^n)^m\implies m=n$

Question

Edit: I goofed up. Need to show that If $$(a^m+b^m)^{n}= (a^n+b^n)^m$$ then $m=n$ where $a,b$ are fixed positive reals and $m,n\in\mathbb{N} $

Answer 1

Factorizing by $a$ you obtain: $$\left(1+\left(\frac{a}{b} \right)^n \right)^{m}=\left(1+\left(\frac{a}{b} \right)^m \right)^{n}$$ with $x=\left(\frac{a}{b} \right)^n$ this is the same as: $$\left(1+x \right)^{m}=\left(1+x^\frac{m}{n}\right)^{n}$$ i.e $$\left(1+x \right)^\frac{m}{n}=\left(1+x^\frac{m}{n}\right)$$ so this is the same as showing that if it exists $x>0$ such that: $$(1+x)^\gamma=(1+x^\gamma)$$ then $\gamma=1$.

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Let $f(x)=(1+x)^\gamma-(1+x^\gamma)$

Then if $\gamma \neq 1$ $$f'(x)=\gamma \left((1+x)^{\gamma-1}- x^{\gamma-1}\right)$$ so if $\gamma >1$ the function is strictly increasing and if $\gamma<1$ strictly decreasing. In both cases as $f(0)=0$ there is no $x>0$ such that $f(x)=0$.

So $\gamma=1$.

Answer 2

Hint: take $m=2,n=3$ then we get $$(a^2+b^2)^3={a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6}$$ $$(a^3+b^3)^2=a^6+2a^3b^3+b^6$$

Answer 3

Counter example:

Take $a = b = m = 1$ and $n = 2$

Then

LHS is $4$ and RHS is $2$.