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What is the problem in my solution of the question: If $\dfrac {a^{n}+ b^{n}}{a^{n-1}+b^{n-1}}=\dfrac{a+b}{2}$,on solving for $n$

One may start this problem by cross multiplying, and I even did the same. But, I am getting stuck at a point where $n$ holds true for every integer value. My attempt: $$ 2\left(a^{n}+b^{n} \right )=\left(a+b\right )\left(a^{n-1}+ b^{n-1} \right )$$ $$ 2a^n + 2b^n=a^n + a\left(b^{n-1}) \right) + b\left(a^{n-1} \right ) +b^n$$ $$ a^n + b^n=a\left(b^{n-1} \right )+ b\left(a^{n-1} \right )$$ $$ a^n + b^n=a\left(\dfrac{b^{n}}{b}\right)+ b\left(\dfrac{a^{n}}{a} \right )$$ $$ a^n + b^n=(b^n)\left(\dfrac{a}{b}\right)+(a^n)\left(\dfrac{b}{a}\right)$$ Now on comparing, we get $$\dfrac{a}{b}=1 \hspace{5 mm} and \hspace{5 mm} \dfrac{b}{a}=1$$

Therefore, 

                                       a=b and b=a

But if this holds true, then $n$ satisfies all values of an integer, which can be seen by putting $a=b$ in the given equation.

The solution for this answer is $1$ and even Other sites are showing their answer for $n$ as $1$, and that holds true by their method. But I am not able to find an error in my method.Plz someone help me in solving this.

P.S-I don’t have a ton of knowledge n mathematics. An answer using elementary math would be appreciated.

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    You are supposed to find $n$ such that $$ a^n + b^n=(b^n)\left(\dfrac{a}{b}\right)+(a^n)\left(\dfrac{b}{a}\right)$$ **for every $(a,b)$** hence the part "Now on comparing" is irrelevant. IOW, one wants that $$ a^n + b^n=(b^n)\left(\dfrac{a}{b}\right)+(a^n)\left(\dfrac{b}{a}\right)$$ even if $$\frac{a}b\ne1\quad\text{and/or}\quad\frac{b}a\ne1$$ – 2017-02-28

1 Answers 1

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One way of explaining is:-

You are asked to find n from the equality. However, coefficients can be compared (and with result deduced subsequently) only when you have an Identity.