Question

Given the integer `N >0` there are exactly `2017` ordered pairs `{x,y }` of positive integers satisfying `1/x+1/y=1/N`. Prove that `N` is a perfect square ?

Answer 1

See below.

Explanation:

`1/x+1/y=1/N rArr N(x+y)=xy rArr (x-N)(y-N)=N^2`

now, `1/x < 1/N rArr N< x` and analogously `N < y` then

`x-N > 0` and `y-N > 0`

At this point we have that the factorizations for `N^2` are in one to one correspondency with the different arrangements for `(x-N)(y-N)` which are `2017`

Now if `p_i` are the prime components of `N` we have

`N = prod_i p_i^(alpha_i)` so `N^2` has

`prod_i (2alpha_i +1)= 2017` factorizations.

But `2017` is prime so `2alpha_1+1=2017` and

`alpha_1 = 2016/2 = 1008` so

`N = p_1^1008 = (p_1^504)^2` which is a perfect square.