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

• MHB
• anemone
In summary, we define two functions, $f(n)$ and $g(n)$, for non-zero integers $n$. $f(n)$ is the sum of all positive integers $b$ for which all solutions $x$ to the equation $x^2+bx+n=0$ are integers, while $g(n)$ is the sum of all positive integers $c$ for which all solutions $x$ to the equation $cx+n=0$ are integers. We are then asked to compute the sum of $f(n)$ and $g(n)$ for all values of $n$ from 1 to 2020.
anemone
Gold Member
MHB
POTW Director
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 $\displaystyle \sum_{n=1}^{2020} (f(n)-g(n))$.

anemone said:
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 $\displaystyle \sum_{n=1}^{2020} (f(n)-g(n))$.
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 $$\sum_{n=1}^{2020} (f(n)-g(n)) = \sum_{s=1}^{44}s = \tfrac12*44*45 = 990.$$

## 1. What does the notation $\sum_{n=1}^{2020}$ mean?

The notation $\sum_{n=1}^{2020}$ represents a summation, where the values of the function $f(n)-g(n)$ are added together for each value of $n$ from 1 to 2020.

## 2. How do you compute the value of $\sum_{n=1}^{2020} (f(n)-g(n))$?

To compute the value of $\sum_{n=1}^{2020} (f(n)-g(n))$, you would need to plug in the values of $n$ from 1 to 2020 into the functions $f(n)$ and $g(n)$, subtract the two values for each $n$, and then add all of the resulting values together.

## 3. Can you provide an example of computing $\sum_{n=1}^{2020} (f(n)-g(n))$?

Sure, let's say $f(n) = 2n$ and $g(n) = n^2$. To compute $\sum_{n=1}^{2020} (f(n)-g(n))$, we would first find the values of $f(n)$ and $g(n)$ for each value of $n$ from 1 to 2020. For example, when $n=1$, $f(n)=2$ and $g(n)=1$, so $f(n)-g(n)=1$. We would repeat this process for each value of $n$ and then add all of the resulting values together, giving us the final sum.

## 4. What is the significance of computing $\sum_{n=1}^{2020} (f(n)-g(n))$?

The value of $\sum_{n=1}^{2020} (f(n)-g(n))$ can provide important information about the relationship between the two functions $f(n)$ and $g(n)$. It can also be used in various mathematical and scientific calculations and analyses.

## 5. Is there a specific method or formula for computing $\sum_{n=1}^{2020} (f(n)-g(n))$?

There are various methods and formulas for computing sums, such as the arithmetic series formula or the telescoping series method. The specific method used to compute $\sum_{n=1}^{2020} (f(n)-g(n))$ would depend on the functions $f(n)$ and $g(n)$ and the desired outcome of the calculation.

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