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

1 Answer
Jun 23, 2017

See below.

Explanation:

1x+1y=1NN(x+y)=xy(xN)(yN)=N2

now, 1x<1NN<x and analogously N<y then

xN>0 and yN>0

At this point we have that the factorizations for N2 are in one to one correspondency with the different arrangements for (xN)(yN) which are 2017

Now if pi are the prime components of N we have

N=ipαii so N2 has

i(2αi+1)=2017 factorizations.

But 2017 is prime so 2α1+1=2017 and

α1=20162=1008 so

N=p10081=(p5041)2 which is a perfect square.