What way would you find numbers with a set number of factors?

[nibble4bits]

If prime numbers are numbers that only factor into 1 and themselves, then what would 3 factors including one and itself be?

It turns out that the middle factor must follow a rule:
1*n^2=n*n=x
The factors of n^2 must be 1, n, n^2. Since n has to be prime, that means
that the answer is the set of all the squares of primes.


Try this for 4, 5, 6, ... factors for x including 1 and x.

[tom11784]

Let Pi be distinct primes for all i:

4 factors: (P1)^3 or (P1)*(P2)
5 factors: (P1)^4
6 factors: (P1)^5 or (P1)^2*(P2)
7 factors: (P1)^6
8 factors: (P1)^7 or (P1)^3*(P2) or (P1)*(P2)*(P3)
so on, so forth...

[nibble4bits]

At 1 and 0 you get the identies 1 and 0. 0 can't be divided at all and 1 has only one possible factor including itself and itself. Hehe we'll just say "period" to make more sense.
At 2 total factors there's only primes. (works both ways: A->B and B->A)
The solutions for 2 factors are a kind of 'key' to the higher-factor-count sets in the 3D tree since obviously primes are the simplest factors possible.
If there's an infinite number of primes, is there an infinite number of the 3-factor results? 4-factor? All? See 2nd post to see why there must be.
Yes, there are infinite primes but no telling how long you'll have to wait to find the next one! I left this in the spoiler so those who want to can do the work themselves to find and understand the proof.

[Numbers]

This is very closely related to something I was thinking about this afternoon.
As there are an infinite number of primes, and as all primes are either 1(mod 6) or 5(mod 6), are there an infinite number of primes 1(mod 6)?

[akruppa]

Yes, this is a special case of the "prime number theorem for arithmetic progressions." Simply put, it says that if you have an arithmetic progression a+b*x with gcd(a,b)=1 and x∈N, you get infinitely many primes. What is more, each such progression for different values of a (but the same b) gets an "equal share" of the primes. See Crandall and Pomerance, Prime Numbers, Theorem 1.1.5.

Alex

[Numbers]

Thank you.

[nibble4bits]

Amazon has one copy of the 2nd edition if you've got $70 (new hard cover math books aren't cheap!) to expand your library. Since I'm near several libraries, colleges and universities, I think I'll be cheap and just go spend some time at a desk in one of them. This should be as interesting as the books by Howard Anton covering some of the more interesting parts of vectors and matrices - assuming you're like me and read that kind of stuff for 'fun'.

ISBN: 0387252827 (there's an older first edition #0387947779 for a little less)

[akruppa]

Some things have been added/changed in the second edition, most notably it now includes the AKS algorithm. But the first edition is still perfectly worthwhile to have and you may be able to get a second hand copy inexpensively now. Maybe check university .market newsgroups or online auctions?

Alex

[Numbers]

Originally Posted by nibble4bits

assuming you're like me and read that kind of stuff for 'fun'.

I do indeed. Last night I read a chapter of "The Art of Calculus" wrapped up in bed with a big mug of cocoa. I spent this morning working through the exercises. This afternoon I did some work on my maths course, and am now relaxing with a chapter of William LeVeque's "Elementary Theory of Numbers". I don't know that it's fun, exactly, but I do find it very rewarding.
Crandall & Pomerance, and of course Knuth were on my wish list for Christmas (again), but sadly Santa saw fit to leave me a mouse and some aftershave. Maybe if I manage to sell another picture before Easter then Amazon will be getting a call.

[ewmayer]

Originally Posted by Numbers

Crandall & Pomerance, and of course Knuth were on my wish list for Christmas (again), but sadly Santa saw fit to leave me a mouse and some aftershave.

You could always feed the mouse to the cat and drink the aftershave...share the joy, I say.

[nibble4bits]

...

Poor cat trying to eat that plastic peripheral.