theorem
Any positive integer n has at most one greater than n \sqrt{n}nThe prime factor of and this is greater than n \sqrt{n}nThe prime factor of is a power of 1.
proof (proof by contradiction)
Proof: at most there is only one greater than n \sqrt{n}nprime factor of
Assume that n exists two larger than n \sqrt{n}nThe prime factors of are p1, p2 respectively.
It is known that p1> n \sqrt{n}n，p2> n \sqrt{n} n
So p1 p2 > ( n ) 2 (\sqrt{n})^2(n)2 = n.
And because n > p1p2
, so simultaneously n > p1*p2 >( n ) 2 (\sqrt{n})^2(n)2 = n
means n>n is contradictory. So the assumption is not true, so at most one is greater thann \sqrt{n}nquality factor of .
Proof: this is greater than n \sqrt{n}nThe power of the prime factor is 1
The following proves that if there is more than n \sqrt{n}nThe prime factor, which should be greater than n \sqrt{n}nThe prime factor is a power of 1.
Assuming that there is one prime factor greater than the root n of n is p, the power of p is k>=2 (not 1)
known p> n \sqrt{n}n, k>=2
so n >= pkp^kpk >= p 2 p^2 p2 > n
so n>n is contradictory.
So the assumption is not true, so the power can only be 1.