Take a prime factor $ p$ of $ a_{1} + 2a_{2} + ... + ma_{m}$, then by Fermat's Little Theorem, $ a_{1}1^{(p - 1)n + 1} + a_{2}2^{(p - 1)n + 1} + ... + a_{m}m^{(p - 1)n + 1}$ is divisible by $ p$ for all $ n$ and as they should be all distinct, we can conclude that there exist infinitely many positive integers n, such that $ a_{1}1^{n} + a_{2}2^{n} + ... + a_{m}m^{n}$ is composite.

KenHungKK wrote: Given an integer $ m\leq$ 2, m positive integers $ a_1,a_2,...a_m$. Prove that there exist infinitely many positive integers n, such that $ a_{1}1^{n} + a_{2}2^{n} + ... + a_{m}1^{m}$ is composite. I think $ m\ge 2$ and $ x_n=a_1*1^n+a_2*2^n+...+a_m*m^n$ composite for infinetely many $ n$. Then $ x_{n+m}=\sigma_1x_{n+m-1}-\sigma_2x_{n+m-2}+...+(-1)^{m-1}\sigma_mx_n$. Were $ \sigma_1=1+2+...+m=\frac{m(m+1)}{2}, \sigma_m=m!, \sigma_k=\sum_{i_1<i_2<...<i_k}i_1i_2...i_k$ Let p is prime, then ${ y_n=(x_n\mod p,x_{n+1}\mod p,...,x_{n+m-1}\mod p})$ is periodic. Therefore we can take $ p=x_k$ and have infinetely many $ n$, suth that $ p|x_n$.

let fix n.let a1*1^n+a2*2^n+....+am*m^n=p where p is prime by little fermat theorem we get p|a1*1^[n+(p-1)k]+a2*2^[n+(p-1)k]+....+am*m^[n+(p-1)k]=b where k integer >0 from this we conclude that b is composite.it's easy to see that problem solved