Require [ \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n} = \dfrac{a+b}{2} ]

[ 2(a^{n+1}+b^{n+1}) = (a+b)(a^n+b^n) ] [ 2a^{n+1}+2b^{n+1} = a^{n+1} + ab^n + a^nb + b^{n+1} ] [ a^{n+1} + b^{n+1} - a^nb - ab^n = 0 ] [ a^n(a-b) - b^n(a-b) = 0 ] [ (a-b)(a^n - b^n) = 0 ]

Since (a \neq b), (a^n = b^n). For distinct positive reals this holds when (n = 0).

Check (n=0): [ \dfrac{a+b}{a^0+b^0} = \dfrac{a+b}{2} ] which is the A.M.

Answer: (n = 0)