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Date: 08/12/2001 at 06:59:55
From: Brian Hayne
Subject: Integers and divisors

Hey Dr Math!

I need some help working out the answer to this question:

Find all composite positive integers M such that the product of 1997 
and M has exactly four divisors.

Date: 08/13/2001 at 19:19:17
From: Doctor Greenie
Subject: Re: Integers and divisors

Hi, Brian -

There is only one.

The number of divisors of a number is directly related to the prime 
factorization of the number. If you are not familiar with this fact, 
try searching the Dr. Math archives from   

using the exact phrase   number of divisors .

Follow one or more of the links that search provides you.

If a number has exactly four divisors, then there are only two 
possibilities for the prime factorization:

(1) The number is the product of two different prime numbers p and q; 
    the divisors are {1, p, q, pq}.

(2) The number is the cube of a prime number p; the divisors are 
    {1, p, p^2, p^3}.

The number 1997 is a prime number. If 1997M has exactly four divisors 
and case (1) above holds, then M must be a prime number other than 
1997. But the problem specifies that M be a composite number. So there 
are no solutions of this type.

If 1997M has exactly four divisors and case (2) above holds, then 
1997M must be the cube of a prime number. Since 1997 is a prime 
number, the only way for 1997M to be a cube is for M to be equal to 

- Doctor Greenie, The Math Forum   
Associated Topics:
High School Number Theory

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