The original question:

*Find…pairs of numbers such that their sum is a factor of their product.*

@BonusQuestion‘s solution is elegantly stated!

”**We can rewrite the equality as (a-k)(b-k) = k^2. Assume c^2 and d^2 are the largest squares that divide a-k and b-k, for two non-negative integers c and d . One can look at the prime factorization of a-k and b-k and notice that: a-k = e c^2 & b-k = e d^2 Therefore k = +/- ecd and hence:**

**a = ec(c+d) [and] b = ed(c+d)**

**OR**

**a = c(c-d) [and] b = ed(d-c)**

**Replace c, d ≥ 0 and e with integers and you get all integer solutions.****”**

Sorry but I’m a little fuzzy on why the same factor e must be shared by (a-k) and (b-k). I agree with the basic equation. Do *both* c^2 and d^2 each divide *both factors* a-k and b-k? Then I think I get it. I’m having trouble reconciling the solution pairs (a,b) which you exemplified earlier by ”(3, 6); (5, 20); (7, 42), (11, 110)” in an earlier post. If the given pair is (n, n*(n-1)) then what are the corresponding values c, d, and e?

I, too, took a course in number theory in high school, which included proving the fundamental theorem of arithmetic, aka unique factorization theorem. Although I generally excelled in math I found number theory (‘queen of mathematics’) to be quite difficult. I guess my brain doesn’t visualize modular congruences that well… ~

When I saw this problem yesterday I soon realized that a and b can’t both be odd integers, reasoning by parity—an even sum won’t divide an odd product. Thus, at least one of the numbers must be even.

This didn’t seem to lead anywhere, however, so I dismissed it as a useless and irrelevant fact. Apparently this was a wise choice…~