Compute $\sum_{n=1}^{2020} (f(n)-g(n))$

[anemone]

For non-zero integers $n$, let $f(n)$ be the sum of all positive integers $b$ for which all solutions $x$ to $x^2+bx+n=0$ are integers and let $g(n)$ be the sum of all positive integers $c$ for which all solutions $x$ to $cx+n=0$ are integers. Compute $\displaystyle \sum_{n=1}^{2020} (f(n)-g(n))$.

 

[Opalg]

For non-zero integers $n$, let $f(n)$ be the sum of all positive integers $b$ for which all solutions $x$ to $x^2+bx+n=0$ are integers and let $g(n)$ be the sum of all positive integers $c$ for which all solutions $x$ to $cx+n=0$ are integers. Compute $\displaystyle \sum_{n=1}^{2020} (f(n)-g(n))$.

Spoiler

The product of the solutions of $x^2 + bx + n = 0$ is $n$. So if both solutions are integers then they must form a factorisation of $n$. Also, if the solution of $cx+n=0$ is an integer then it must be a factor of $n$. So the ingredients of the sums $f(n)$ and $g(n)$ must all arise from factorisations of $n$.

If $n=st$ is a factorisation of $n$ as a product of two positive integers then the solutions of $x^2 + (s+t)x + n = 0$ are $x= -s$ and $x = -t$. So $b=s+t$ is one of the values of $b$ that go into the sum $f(n)$. Also, $x=-t$ is the solution of $sx + n = 0$ and $x=-s$ is the solution of $tx+n=0$. So $s$ and $t$ are two of the values of $c$ that go into the sum $g(n)$. Therefore the net contribution of the factorisation $n=st$ to $f(n) - g(n)$ is $(s+t) - (s+t) = 0$. Summing over all possible factorisations of $n$, you see that $f(n) - g(n) = 0$.

But that argument goes wrong if $n$ is a perfect square, because the factorisation $n = s^2$ gives a contribution $2s$ to $f(n)$, but it only contributes $s$ to $g(n)$. Therefore $f(s^2) - g(s^2) = s$.

The largest square that is less than $2020$ is $44^2 = 1996$. Therefore $$ \sum_{n=1}^{2020} (f(n)-g(n)) = \sum_{s=1}^{44}s = \tfrac12*44*45 = 990.$$