Season 7 Episode 1

Take a large power of two and calculate two to the power of that. Is the answer one less than a prime? In this episode; probably not!

Further Reading

Fermat primes

On the stream I did something silly and considered the sequence $2^{2^n}-1$. That’s $F_n-2$ and that’s also the difference of two squares, so it factorises as $(2^{2^{n-1}}+1)(2^{2^{n-1}}-1)$

That’s disappointing if I’m looking for primes, but we can salvage something interesting from this factorisation! Let’s recognise that $2^{2^{n-1}}+1$ is $F_{n-1}$ and also that $2^{2^{n-1}}-1$ is $F_{n-1}-2$. But hang on, that’s the same sort of thing that we started with. So we can keep going, replacing each $F_{n-1}-2$ with a factorisation of that number.

If you continue this pattern, you should be able to prove that $F_n=F_0\times F_1\times F_2\times\dots\times F_{n-2}\times F_{n-1}+2$. What can you deduce about the prime factors of Fermat numbers?

See also; Sylvester’s sequence!

Leo showed us that if $n$ has an odd factor then $2^n+1$ factorises. Test your understanding; if $m$ is not a prime number, what can you say about $2^m-1$? 

See also; Mersenne primes! 

 

 

Prime Number Theorem 

The fourth year of the Mathematics degree at Oxford consists of many optional courses, one of which is C3.8 Analytic Number Theory, currently taught by James Maynard. We could jump right in and look at the proof (the lecture notes are online), but the prerequisites for the course are "Basic ideas of complex analysis. Elementary number theory. Some familiarity with Fourier series will be helpful but not essential." so we might have hard time. 

In Oxford, Complex Analysis is taught in the second year and Fourier Analysis is taught in the third year. In that sense, parts of the course are cumulative, building on previous knowledge.

Here's a flavour. Complex Analysis looks at the properties of functions that take complex number input and return complex number output. There are some really cool theorems in complex analysis relating properties of the function to its behaviour at certain points. The idea in the proof is to apply this sort of technology to the Riemann zeta function. You might have heard of that function; there's a famous unsolved problem asking for the locations of the zeros of the Riemann zeta function. Our understanding of the Riemann zeta function is closely related to our understanding of prime numbers. Quanta magazine have made a nice video explainer on this.

 

Fermat’s Little Theorem 

Eagle-eyed people who were here last week will have noticed that Jonah proved $a^{p}\equiv a \,(\text{mod} \,p)$ but this week I wanted to use $a^{p-1}\equiv 1 \,(\text{mod} \,p)$. We can get to the first to the second if we can "divide by $a$". Why might we not be able to divide by $a$? Here’s an example of what can go wrong; $2\times 8 \equiv 2\times 3 \,(\text{mod}\, p)$ but $8$ and $3$ are not equivalent modulo 10.

The key idea is that any number $a$ has a "multiplicative inverse" modulo $p$ if $p$ is a prime number. The proof of this is rather nice, and it might appear on the Maths Club one day (look up Bezout’s Identity if you can't wait, or read this NRICH page for a half-way step to finding such an inverse). The lesson for now is "if someone shows you a new sort multiplication, do not assume that you can divide". The first rule of maths club is "do not divide by zero", but maybe that should be "do not divide by anything that doesn’t have a multiplicative inverse"! 

 

Sophie Germain primes 

The chains of Sophie Germain primes we were investigating at the end of the stream are called Cunningham chains.  Allan J C Cunningham was a British-Indian mathematician working at the start of the 20th century. At time of writing, the longest known chain has 17 primes.

 

If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk. 

Please contact us with feedback and comments about this page. Last updated on 12 Jan 2024 16:31.