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### 1997M

```
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

http://mathforum.org/mathgrepform.html

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
1997^2.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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