# Any neat proof that $0$ is the unique solution of the equation $4^x+9^x+25^x=6^x+10^x+15^x$?

It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strict monotonic increasing functions. It is also easy to check that 0 is a solution of the equation. Also I chart the functions, and it looks that for any $x$, $f(x)>g(x)$, which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ it's an absolute minimum point for $h(x)$. However $h(x)$ it is a function with a messy derivative, and is not looking easy(for me) to find the zeroes of this derivative.

Does anyone, know an elegant proof(maybe an elementary one, without derivatives) for this problem?

## 3 Answers

HINT: $$a^2+b^2+c^2\geq ab+bc+ca$$

• Thank you very much, I got it! – motoras Aug 18 '17 at 19:22
• ok, that is nice! – Dr. Sonnhard Graubner Aug 18 '17 at 19:23
• Very nice observation! – Cauchy Aug 18 '17 at 19:23
• From $(a-b)^2+(b-c)^2+(c-a)^2\ge 0$ – Mark Bennet Aug 18 '17 at 19:25
• at the University i gave a talk about "solving equations with inequalities" – Dr. Sonnhard Graubner Aug 18 '17 at 19:25

Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x \geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"

I guess you want a proof that works for the continuous case. If we would know that $$x$$ is always an integer, then we could say that $$f(x)$$ is an even number for any $$x$$ integer except $$0$$, and $$g(x)$$ is odd for any $$x$$ integer except $$0$$. So, $$x$$ must be zero.