Are there infinite numbers with four or six factors?

By Ali Korotana (P4195) on Monday, May 7, 2001 - 08:07 pm :

There exists a way of showing that there are infinite numbers with two factors, i.e. prime numbers. Is there a way of showing that there are infinite numbers with four or six or however many factors? I ask since we are looking for values of n where (n!)² -1 has four factors.

By Dan Goodman (Dfmg2) on Monday, May 7, 2001 - 08:17 pm :

Sure, if p₁ =2, p₂ =3, ... are the prime numbers then p_i p_j (where i is not equal to j) has 4 factors: 1, p_i , p_j , p_i p_j . If you want 6 factors, try p_i ² p_j (where i is not equal to j), since the factors are 1, p_i , p_i ² , p_i p_j , p_j , p_i ² p_j . And so on...

By Ali Korotana (P4195) on Monday, May 7, 2001 - 08:28 pm :

Thats quite interesting but I'm not sure I'm really understanding why it 'works', could you elaborate a bit please Dan?

By Dan Goodman (Dfmg2) on Monday, May 7, 2001 - 10:01 pm :

Ali, do you know about unique factorisation? Every integer can be written uniquely as a product of prime numbers. Once you've got this, it's reasonably easy to see that the factors of, say,

p_{i}^{2} p_{j}

are what I said and that there are no others. In fact, there are an infinity of numbers with

n

factors if

n > = 2

. See if you can extend the method I used for 4 and 6 factors to work for

n

factors by writing

n

as a product of primes. It's quite tricky so if you can't work it out after a bit I'll post the solution. The key fact you need to use is that all of the factors of

p_{1}^{n_{1}} p_{2}^{n_{2}} \dots p_{k}^{n_{k}}

are just the numbers

p_{1}^{m_{1}} p_{2}^{m_{2}} \dots p_{k}^{m_{k}}

where each

m_{i} \leq n_{i}

By Dan Goodman (Dfmg2) on Monday, May 7, 2001 - 10:42 pm :

D'oh! Just realised that there's a very easy proof of the problem I set above. Still, I'll leave it unsolved to see if you can get it.

To make it a bit more difficult for you, try and prove that if $n = p_{1}^{n_{1}} \dots p_{k}^{n_{k}}$ then there is an infinity of numbers $r$ with $n$ factors such that the number of prime factors of $r$ is $n_{1} + \dots + n_{k}$ . Hence prove that there is an infinity of numbers $r$ with $n$ factors such that the number of prime factors of $r$ is $\geq \log_{2} n$ .