# Comparing exponents in an equation.

1. Nov 2, 2013

### AakashPandita

an + bn/ an-1 + bn-1 = (a+b)/2
can we say n=1 by comparing exponents?

is there any other solution of it?

Last edited: Nov 2, 2013
2. Nov 2, 2013

### Simon Bridge

$$\frac{a^n + b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$$ rearranging: $$a^n+b^n = ab^{n-1} + ba^{n-1}$$

By inspection, n=1 will satisfy the relation for all real a and b.
Are there any other examples?

$n=2$

$a^2+b^2=2ab \Rightarrow (a-b)^2=0$
... so the relation with n=2 describes a curve (a parabola).

So it kinda depends on what counts as an answer.
(The problem is under-specified.)