2019!

13^2 +37^2+13*37=2019

$x^2 + xy +y^2 = (x+y)^2 - xy$ This fact means we can look for squares a little larger than 2019, and see if the difference can be factored into two numbers, such that those numbers sum to the number you’re squaring.

Since that sentence is really confusing, here’s an example: The first square number greater than 2019 is 2025, or 45 squared. Our difference here is just 6, and we need to factor that into two numbers who sum to 45. Obviously we can’t, so 45 isn’t the solution.

If you keep trying larger squares, you’ll arrive at 50, since $2500-2019=481=13*37 , 13+37=50$ Obviously this works, but after initially solving the problem, I thought of a potentially ‘better’ way. We are looking for a pair of numbers which sum to one quantity, and multiply to another. This is also occuring in a quadratic equation. I claim that the answer to this problem is the number, k, such that the roots of

$z^2 + kz + (k^2-2019)$

are both integers (x and y).

I put better in quotes for a reason. This approach is more complicated and confusing than the trying numbers approach, and likely takes longer in practice. The only reason I’d say is better is that it’s more general, which is usually the gold standard for math problems. I guess my point is this: if you find it easier to bash a problem some, go for it! I feel like this approach is usually downplayed, but it can be extremely effective. Sometimes, it will lead you to the general solution, like it did for me here. Point is, solve how you want to solve, and have a nice day. :)

If you’ve just scrolled down to find the solution:

$13^2 + 13*37 + 37^2 = 2019$

so the answer is $13+37=\boxed{50}$