Comparing exponents in an equation.

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The discussion centers on the equation \( \frac{a^n + b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2} \) and explores the implications of setting \( n=1 \). It is established that \( n=1 \) satisfies the equation for all real values of \( a \) and \( b \). Additionally, the case for \( n=2 \) is examined, revealing that it leads to the equation \( a^2+b^2=2ab \), which simplifies to \( (a-b)^2=0 \), indicating a parabolic relationship. The discussion concludes that the problem is under-specified, leaving room for interpretation regarding additional solutions.

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

is there any other solution of it?
 
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$$ \frac{a^n + b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$$ 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.)
 

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