Determine how many integer ordered pairs $(x, y)$ satisfy the equation $\mid x + 2020\mid + \mid y + 505\mid = 4$.

Question

I do not see how this is possible if all of it equals to a number higher than $4$. If $x$ or $y$ is negative would it make a difference? How can $x$ and $y$ adding two big numbers equal $4$. Is there a faster way to figure out the amount of pairs?

Answer 1

There are 5 ways to write 4 as a sum of two non-negative integers. We are only concerned about non-negative integers because $|x + 2020|$ and $|y + 505|$ can only be non-negative integers if $x, y \in \mathbb{Z}$

$4 = 0 + 4 = 1 + 3 = 2 + 2 = 3 + 1 = 4 + 0$

So, for each pair, we are going to get values of x and y.

Like for (0, 4), $|x + 2020| = 0$ and $|y + 505| = 4$, we got to solve these two equations.

For $|x + 2020| = 0$, we get $x = -2020$

And, for $|y + 505| = 4$, we get $y = -501, -509$

So, for pair (0, 4), we get $(x, y) = (-2020, -501), (-2020, -509)$.

Now I think you can do the same with the last 4 pairs and you would get the full solution. By the way, if you want to know just the number of solutions, then we can observe that when $x = 0$ or $y = 0$ we get only one value of $x$ or $y$. In any other value, we get 2 distinct values. For pair (0, 4) and (4, 0) there's going to be 2 solutions for each. But for pair (1, 3), (2, 2) and (3, 1), we will get 4 solutions for each. So we get a total of $2 * 2 + 4 * 3 = 16$ solutions.