If $a+b = 2020$ and $ab$ is a multiple of $2020$ , find all pairs of $(a,b)$

Question

If $a+b = 2020$ and $ab$ is a multiple of $2020$ , find all pairs of $(a,b ) $ for $a,b \in \mathbb N.$

My take : $$2020a-a^2 = 2020k$$ $$a = \dfrac{2020 \pm \sqrt{2020^2 -4\cdot2020k}}{2}$$

Since $a$ is an integer , $\sqrt{2020^2 -4\cdot2020k}$ must be an integer . $$2020^2 - 4\cdot2020k = x^2\implies 2020\,(\,2020 - 4k) = x^2$$

But this doesn't seem lead me anywhere . Is there a better way to solve this problem ?

Answer 1

We have :

$$a+b = 2020 \implies b = 2020-a$$ $$ab = a(2020-a) \implies 2020a-a^2 \equiv 0\mod2020$$

We conclude that $a^2$ must be be divisible by $2020$ , or :

$$a^2 = 2020\lambda \implies a = \sqrt{2020\lambda}$$

From the fact that $2020 = 2^2\cdot5\cdot101$ , we must have $2020\lambda$ to be of the form $\color{#d01}{\lambda = 5\cdot101\cdot k^2}$ or $\color{#3d1}{\lambda =2^2\cdot5\cdot101\cdot k^2 }$ .

The reason is to mulitply $2020$ by a number so as to make it a perfect square . Note that other representations of $\lambda$ will make $a\gt2020$ which is not possible.

So $$a =1010k_1 \quad \text {or} \quad a = 2020k_2$$

where $K_1 \in\{0,1,2\}$ and $K_2 \in\{0,1\}$.

Inputting the values of $k_1$ and $k_2$ , we get the foll0wing unique pairs : $$(a,b) = \begin{cases}\color{#4af}{(0,2020) \\(1010,1010) \\(2020,0)}\end{cases}$$