Conjecture About De Koninck

For a positive integer n, let ?(n) and ?(n) denote the sum of divi-sors of n and the product of distinct prime divisors of n, respectively. Here ?(1)=1. In 2000, De Koninck conjectured that ?(n)=?(n)2 has only two solutions n=1 and n-1782.In 2012, Broughan and De Koninck proved that if n>1 and ?(n)=?(n)2, then the prime factorization of n has the form are even and either p?p1??1?1 (mod 4) or p(?)3 (mod 8) and ?1 is even. In 2014, Broughan etc. proved in J. Number Theory that n is divisible by the forth power of at least one odd prime.A positive integer n is called a De Koninck's number if ?(n)= ?(n)2. In this thesis, we make further research on De Koninck num-bers. We prove that if n is a De Koninck number and n?1,1782 such that only one exponent of odd primes is equal to 1 in the prime factorization of n, then(1) n is divisible by 3;(2) n is divisible by the forth powers of at least two odd primes;(3) if p | n and the exponent of p is equal to 1 in the prime factorization of n, then p> 1571 and at most two different odd prime divisors of n are greater than p;(4) at least two exponents of odd prime divisors in the prime factorization of n are equal to 2.The results are published in Journal of Number Theory 154 (2015), 324-364.