# Open problem in number theory

3 Dec

A number theorist was trying to convince me that while you can do things easily for finite sets of prime numbers, it’s really hard to make a leap to infinitely many primes.  She gave me a sketch of an example, and I thought I’d share it/some other interesting number theory things.

We say that an integer is squarefree if for every prime number p, $p^2$ does not divide a.  Remember, the primes are the numbers that are only divisible by themselves and 1, like 2, 3, 5, 7, etc.  Every integer can be written as a product of prime numbers, so squarefree means what we think it does: no perfect square factors.  Here are the first squarefree numbers which aren’t primes: 6, 10, 14, 22, 26, 30,…

So you can ask a few questions as soon as you get this definition.  How many squarefree numbers are there?  A: infinitely many, because any product of two primes is squarefree, and there are infinitely many primes.  But how big is that infinity?  Remember, not all infinities are equal.  Maybe a better question is, what’s the ratio of squarefree numbers vs. not-squarefree numbers?  This seems a little nuts because we’ll have infinity divided by infinity, but it’s not nuts to say that half the numbers are even.  So we’re not nuts!  Huzzah! This is from wolfram alpha and shows 10000 numbers: the white pixels are squarefree. 1-100 are the bottom horizontal line of pixels, you can see 1, 4, 8,9,12,16,18,20 all blacked out in the lower left.

A: about $6/\pi^2$ much of the numbers are squarefree, and this has been known since at least 1951.  Just from the first 20, we have 60% squarefree, and $6/\pi^2$ means that as you look at bigger and bigger numbers, the proportion of squarefree numbers approaches 60.97….%  But this is crazy, right?  Where did that $\pi$ come from?

Well, you can go prime by prime.  For instance, about 1/9 of numbers are divisible by 9, so 8/9 of numbers are 9-free.  And 3/4 are 4-free.  So 3/4*8/9=2/3 are 4-and-9-free.  You could do this for any finite set of primes, and say that the chance that is $latex p^2$-free for all of your finite set is $\prod_{p\in S} (1- \frac{1}{p^2})$.  Then we’d conjecture that you could make this an infinite product $\prod_{primes} (1-\frac{1}{p^2})$.

By this crazy formula from 1737 (oh that Euler!), $\sum_1^{\infty} n^{-s} = \prod_{primes} \frac{1}{1-p^{-s}}$.  And then that sum on the left is the definition of the Riemann zeta function, and $\zeta(2) = \frac{\pi^2}{6}$, which was proven by Euler 91 years after Basel posed the question.

But that step where we conjectured that you could jump to an infinite product?  It works in this case, but through a lot of hard work!  Which I do not understand nor will explain.

Here’s a related question.  Suppose you have a polynomial (a function in the form $f(x) = a_nx^n + a_{n-1}x^{n-1}+\ldots+ a_1x + a_0$), and you put numbers into it.  How many of those resulting function values are squarefree?  If you have something silly like $f(x) = 9x+9$, all values will be divisible by a square.  So let’s cut out polynomials like that.  Do squarefree values happen infinitely often?  And if so, how often [in the infinite sense that half of numbers are even]?

Open question: Does $f(x) = x^4+2$ take infinitely many square-free values?  If so, what is the probability that f(x) is squarefree, if you randomly choose an x?

You can ask a question with just putting in primes instead of putting in all numbers for too.  A polynomial is irreducible if you can’t write it as a product of other polynomials.  For instance, $x^3-1$ is not irreducible, since $x^3-1 = (x-1)(x^2+x+1)$, but $x^2+x+1$ is irreducible.  The degree of a polynomial is the biggest number that appears as an exponent of x, so the degree of $x^3-1$ is 3.

Open question (with conjecture): If f(x) is an irreducible polynomial of degree 3 or more, how many squarefree values does it take?

Dude, number theory is full of unsolved problems that are easy to state!  When reading up for this post, I ran into this magic squares problem.  Remember a magic square is one where the sum of all the numbers in each column, in each row, and along the diagonals is all the same number. Sum of rows, columns, and diagonals is all 15.  I’m sorry I didn’t bother to make you a magic square; these values are from wikipedia.

Open question: Build a magic 3×3 square of square numbers.

I hope I’ve managed to convey some interesting number theory to you today!  Personally I’m not that into number theory, but I tried to not let that color this post.

Look at this bibimbap I made for dinner the other day!  My first time making bibimbap.

A photo posted by Yen Duong (@yenergyyyy) on Dec 1, 2015 at 7:18am PST

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