[Peter]

Determine all integers such that is an integer.

[Rust]

Obviosly n is odd. Let is maximal prime divisor n, suth that and , then .
If , then must be .
Therefore from all solution n we get and find minimal solution (n>1) , Because had not another prime factors. And all solutions are and .

[Peter]

Rust wrote:

Let is maximal prime divisor n, suth that and , then .

Who says there is such a prime divisor?

Rust wrote:

If , then must be .

Why?

[chien than]

http://www.kalva.demon.co.uk/imo/isoln/isoln903.html

[kimnimalar]

Rust wrote:

Obviosly n is odd. Let is maximal prime divisor n, suth that and , then .
If , then must be .
Therefore from all solution n we get and find minimal solution (n>1) , Because had not another prime factors. And all solutions are and .

please write full, I can't understand vp(m) and others... what's mean this function?

[ZetaX]

For notations, always take a look at http://www.artofproblemsolving.com/viewtopic.php?t=76610 .

[Rust]

kimnimalar wrote:

Rust wrote:

Obviosly n is odd. Let is maximal prime divisor n, suth that and , then .
If , then must be .
Therefore from all solution n we get and find minimal solution (n>1) , Because had not another prime factors. And all solutions are and .

please write full, I can't understand vp(m) and others... what's mean this function?

Full solution:
Because for is odd, n is odd. Let p is maximal prime divisor of n and . Then and . If obviosly . Let .Therefore exist odd u ( - even if m>1 and odd), suth that .
It give . Because , we get . It give new solution m.
If , then , therefore .
Because , we get . Exactly .
From solution we get solution , suth that and had less prime divisors, then .
From we get ,... and . Therefore .
It give solution . If , then . Therefore we had not another solutions.