Hmm I hope this is correct. Assume otherwise. The condition rewrites as $(a_{n+2}+a_{n+1})^m=2^m(a_{n+1}^{m-1}+a_n^{m-1})$. We claim that $a_i<a_1$ for all odd indices $i$. Indeed, $(a_3+a_2)^m=2^m(a_2^{m-1}+a_1^{m-1}) \leq a_1(a_2^{m-1}+a_1^{m-1})<(a_1+a_2)^m$, and so $a_3 <a_1$. Now, assuming that $a_{2N-1}<a_1$, we obtain $(a_{2N+1}+a_{2N})^m=2^m(a_{2N}^{m-1}+a_{2N-1}^{m-1})<a_1(a_{2N}^{m-1}+a_1^{m-1})<(a_1+a_{2N})^m$, hence $a_{2N+1}<a_1$, as desired. Therefore, $a_i<a_1$ for all odd indices $i$, and since all $a_i$ are integers, we conclude that there are infinitely many odd $i$ such that $a_i=K$ and $a_{i+2}=M$ for some fixed $K,L$. Note that $a_{i+1}$ can attain at most $m$ values, so we infer that there are infinitely many odd $i$ such that $a_{i}=K,a_{i+1}=L,a_{i+2}=M$ for some fixed $M$, too. Hence, tracing back we obtain that the sequence $(a_n)$ is periodic, with even period, let it be $T$. Then, $T-1$ is odd, and so $(a_T+a_1)^m=2^m(a_{T-1}^{m-1}+a_T^{m-1})<a_1(a_1^{m-1}+a_T^{m-1})<(a_1+a_T)^m$, which is a contradiction.

Orestis_Lignos wrote: Hence, tracing back we obtain that the sequence $(a_n)$ is periodic, with even period, let it be $T$. Sorry, but why is $T$ even? Actually, I think you could end here with $a_1=a_{2T+1}<a_1$.

seems similar to a bulgarian problem. So taking motivation from it , can we use riemann sums to prove this ( the use of integration in the bulgarian problem was discussed by dgrozev in his blog!)