Question

pe

Answer 1

Let's try and wrap our heads around what the statement is saying. What happens when x is small?

Take for example, x=1, 2, and 3.

What do we find f(1), f(2), and f(3) to be?

Answer 2

@Luis_Rivera How did you get that? Could you provide a proof of your statement for others to see?

Answer 3

1a+1b=1x
a+bab=1x
ab=(a+b)x
x2=ab−(a+b)x+x2=(a−x)(b−x)

Therefore, the pair (a,b) is a solution if (a−x)(b−x) is a factorization of x2 with a−x and b−x both larger than −x.

It is specified that x=pn. The factorizations of x^2=p^ 2n are all of the form p^k*p^2n−k, with 0≤k≤2n, hence there are 2n+1 factorizations and thus 2n+1 solutions to the original equation. Therefore, 2n+1=13, and n=6.
The smallest number x of the form p6 is x=2^6=64

Answer 4

Beautiful proof! : )