For all $n\in\Bbb N$ does there exist $n_1$ and $n_2$ such that $d(n_1) +d(n_2)=n$

Question

I am trying to show that for each positive integer, $n$, we can find integers $n_1,n_2$ that satisfy $$d(n_1) +d(n_2)=n$$ where $d(n)$ is the divisor function. I am frustrated because even with easy number theory problems like this I am unable to see what to do or how to approach it. I know that if $$n_1=p_1^{\alpha_1}p_2^{\alpha_2}...p_s^{\alpha_s}$$ where $p_i$ are distinct primes, then $$d(n_1)=(\alpha_1+1)(\alpha_2+1)...(\alpha_s+1)$$ but I don't see how to expand this and add $d(n_2)$ to get the positive integer $n$. Any help is appreciated.

Answer 1

Let $n\in\Bbb N$. Then if $n=1$ this is impossible since $d(k)\ge 1$, so $d(n_1)+d(n_2)\ge 2$, however for $n\ge 2$ let $n_1=2^{n-2}$ and $n_2=1$. Then $d(n_1)=(n-2+1)=n-1$ and $d(n_2)=1$, so we have

$$d(n_1)+d(n_2)=n-1+1=n$$

as desired.