Lecture 10: Prime numbers
Julia Collins
27th November 2013
Unsolved prime number problems
Given that prime numbers are the building blocks of all numbers, it is remarkable that there
are so many simple problems we haven’t got answers to yet.
1. The Goldbach conjecture: Every even number greater than 2 is the sum of two
prime numbers. For example, 6 = 3 + 3, 12 = 5 + 7, 100 = 47 + 53. This was proposed
in 1742 by Christian Goldbach, and has now been verified for numbers up to 4 × 1018 .
A recent paper (May 2013) by Harald Helfgott has shown that every odd number
greater than 5 is the sum of at most three primes: http://www.truthiscool.com/
prime-numbers-the-271-year-old-puzzle-resolved.
2. The twin prime conjecture: There are infinitely pairs of primes with a difference
of two. For example, some twin prime pairs are (11,13), (17,19), (41,43). The largest
known twin primes have over 200,000 digits in them – but can we always find a bigger
one? More generally, Polignac’s conjecture states that for any even number n there
are infinitely pairs of primes separated by n. This has not been proven or disproven
for any value of n. [Trivia: primes separated by six, for example (5,11) or (13,19),
are called sexy primes.] However, a paper by Yitang Zhang in April 2013 was the
first to show that the conjecture is true for some value of n less than 70 million.
http://goo.gl/YK6sF A longer article about progress on this is in the Links section
of the Open Studies website.
3. The Fermat prime conjecture: There are infinitely many Fermat primes. A Fermat
n
1
prime is a prime of the form 22 + 1; for example 5 = 22 + 1. Fermat conjectured
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that all numbers of this form were prime, but in 1735 Euler showed that 22 + 1 was a
composite number. Nobody has found a Fermat prime other than up to n = 4, but it
is unknown whether there might be infinitely many.
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4. The Mersenne prime conjecture: There are infinitely many Mersenne primes. A
Mersenne prime is a prime of the form 2n − 1; for example 7 = 23 − 1. A Mersenne
number can only be prime if n itself is prime. Forty-eight such Mersenne primes are
known (as of May 2013), including the largest prime ever discovered (in January 2013),
but it is unknown whether there are infinitely many such numbers.
5. Quadratic primes: Is there a quadratic polynomial which takes prime values infinitely
often? Euler discovered the quadratic polynomial n2 − n + 41 which generates prime
numbers for n = 0, ..., 40. But does it keep on generating new primes as we keep trying
new values of n? (We know that not *every* value of a quadratic can ever be prime,
but does every polynomial ‘run out’ of prime values or is there one which can keep
generating primes?) There does seem to be something special about the relationship
between quadratic polynomials and primes, as we can see on the Ulam spiral where
primes tend to like congregating on certain diagonals.
6. The Riemann Hypothesis: Every (non-trivial) zero of the Riemann zeta function
has real part equal to 12 . This is one of the six remaining Clay Institute Millennium Prize
Problems worth $1 million. It is particularly intriguing to prime-number enthusiasts
because the zeros of the zeta function dictate the distribution of the prime numbers.
The Riemann zeta function
There are two different ways to write down what the zeta function is.
1
1
1
1
1
+ s + s + s + s + ...
s
1
2
3
4
5
(1)
1
1
1
1
1
=
× ...
1 ×
1 ×
1 ×
−s
1−p
1 − 2s
1 − 3s
1 − 5s
1 − 71s
p prime
(2)
ζ(s) =
and
ζ(s) =
Y
Why are these the same? It seems incredible that an infinite product with prime numbers
could be same as an infinite sum of integers. In particular, it means that the fact that there
are an infinite number of primes is a direct consequence of the harmonic series (1 + 12 + 31 +
1
+ . . . ) being infinite. (Put s = 1 into both equations!)
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To show that both equations are the same thing, we need to use the formula for the sum
of a geometric series and the Fundamental Theorem of Algebra...
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Proving the equivalence of the two forms of the zeta function
If we have a geometric series, for example 1, 12 , 14 , 18 , . . . where the terms are being multiplied
by a constant factor, then we can write down a formula for its sum:
1 + x + x2 + x3 + x3 + · · · =
1
1−x
for |x| < 1
This looks a bit like the prime terms in formula (2).
This means we can write
1
1
1
1
1
=
1
+
+
+
+
+ ... ,
1 − 2−s
2s 22s 23s 24s
1
1
1
1
1
=
1
+
+
+
+
+ . . . etc
1 − 3−s
3s 32s 33s 34s
Now we can write ζ(s) as a product of infinite series:
1
1
1
1
1
1
1
1
1
ζ(s) = 1 + s + 2s + 3s + . . .
1 + s + 2s + 3s + . . .
1 + s + 2s + 3s + . . . . . .
2
2
2
3
3
3
5
5
5
What does this infinite number of infinite brackets look like when multiplied out? We
will get every possible combination of powers of primes appearing to the power of s in the
denominator. Every number of the form ( 21a 31b 51c 71d . . . )s will appear, for whatever choice
you’d like of a, b, c, d, . . . .
The Fundamental Theorem of Algebra tells us that every number can be written uniquely
as a product of primes; for example 60 = 22 × 3 × 5. This means that 601s will appear as one
of the terms in the zeta function because it’s the product of 212s , 31s and 51s . Using the same
argument, every number of the form n1s is a term of the zeta function. This is what gives us
formula (1) for the zeta function.
[You can see another proof of the equivalence on Wikipedia: http://en.wikipedia.
org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function.]
Modular arithmetic
Modular arithmetic is a way of working with numbers where we only care about remainders.
You are already familiar with this: when we ask what the time is 3 hours after 11 o’clock,
we would say 2 o’clock, not 14 o’clock (or 26 o’clock). In a clock system, we assume that
there are only 12 numbers and that every time you get back to 12 o’clock you start again at
0. So 13 = 1, 14 = 2, 29 = 5 and so on.
When mathematicians do clock arithmetic, they say they are working modulo 12, or mod
12. If you want to know what a number is mod 12, you look at the remainder of the number
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after you divide by 12. For example, 29 mod 12 = 5, because 29 = (2 × 12) + 5. But there
no reason why 12 is special: we can work modulo whatever number we want! If we work
modulo 9, then 29 mod 9 = 2 because 29 = (3 × 9) + 2. Just take out as many multiples of
the number (in this case, 9) as you can, and see what’s left over.
Modular arithmetic may seem like a strange thing to do, but it makes mathematics much
easier. Instead of infinitely many numbers in the number line, there are only finitely many
in this universe. With clock arithmetic, we only have to know 12 numbers. And we can do
most of the usual things we can do with normal numbers: add, subtract, multiply (but not
always divide!). If you want, you can practice your modular arithmetic skills on this website:
http://www.math.csusb.edu/notes/cgi/modtutor.cgi.
[Modular arithmetic is a massive and beautiful subject in its own right, but for now this
is all you need to understand the stuff about mind reading and cryptography!]
Mathematical mind reading - guess the digit
Method: Ask someone to write down a 6-digit number, in which the digits aren’t all the
same (so not 111111). Then ask them to write down the same 6 digits again but in a different
order. Now they should subtract the smaller number from the larger one. In the answer,
they should read out to you all of the digits except one, in any order, and you will tell them
the digit they’ve missed out. (To make 100% sure you’ll guess correctly, tell them they
shouldn’t pick a zero as the missing digit.)
The trick: To understand how this trick works, all you need to know is that the value of
a number modulo 9 (i.e. the remainder on division by 9) is the same as the value of the sum
of its digits modulo 9. To see this, take the example of 247 and let’s think about breaking
it down into multiples of 9:
247 = (2×100)+(4×10)+7 = (2×(99+1))+(4×(9+1))+7 = (2×99)+(4×9)+(2+4+7).
So if we take out all the multiples of 9 from 247, the remainder we get is the same as the
remainder if we just took 9s out from 2+4+7. And this is true for ANY number written in
base 10, because 10n always has remainder 1 when we divide by 9.
In particular, notice that a number is divisible by 9 whenever the sum of its digits is
divisible by 9.
In the two six-digit numbers your partner first writes down, the sum of the digits is the
same for each because they’re just the same numbers in a different order. So these two big
numbers have the same value modulo 9.This means that if we subtract one from the other,
the answer must be 0 (modulo 9) which is the same as saying it is divisible by 9. (Example:
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if one number is a multiple of 9 plus 5, and another number is a different multiple of 9 plus
5, then when we subtract them the 5s will cancel and the answer will just be a multiple of
9.)
Since the answer is a multiple of 9, the sum of its digits is a multiple of 9. To discover
the missing digit, then, add up the numbers they read out to you and tell them the number
which would make this sum into the next multiple of 9. For example, if they read out 1,4,6,3,
then this adds up to 14 and you need to add another 4 to get to the next multiple of 9, so 4
is the missing digit.
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