China Western Mathematical Olympiad

For any given positive integer $k$, choose positive integer $m$ sufficiently large such that $2^m+3^m-k>1$. Consider the following $k$ integers: $$2^m+3^m-1,2^m+3^m-2,\dots ,2^m+3^m-k,$$ all of which are larger than $1$. From each of these integers, pick a prime factor: $p_1,p_2,\dots ,p_k$, and let $$n_t=m+t(p_1-1)(p_2-1)\cdots (p_k-1),$$ where $t$ is an arbitrary positive integer. For any fixed integer $i$ ($1\le i\le k$), one has $2^{n_t}\equiv 2^m(\mathrm{mod}{p}_i)$. In fact, if $p_i=2$, then the result is obvious. Assuming that $p_i\ne 2$, it follows from Fermat's little theorem that $$2^{n_t}=2^m\cdot 2^{t(p_1-1)(p_2-1)\cdots (p_k-1)}\equiv 2^m\cdot 1=2^m(\mathrm{mod}{p}_i).$$ Similarly, one has $3^{n_t}\equiv 3^m(\mathrm{mod}{p}_i)$. Observe that $$2^{n_t}+3^{n_t}-i\equiv 2^m+3^m-i\equiv 0(\mathrm{mod}{p}_i),$$ $$2^{n_t}+3^{n_t}-i>2^m+3^m-i.$$ Hence, $2^{n_t}+3^{n_t}-i$ is a composite number. Therefore, $n_t$ is one of the positive integers $n$ such that $$2^n+3^n-1,2^n+3^n-2,\dots ,2^n+3^n-k$$ are all composite. As $t$ is arbitrarily chosen, there are infinitely many such positive integers satisfying the conditions above.