Question 1

What is the largest prime that you can find with decimal expansion \(2015\cdots\). Your answer should include a verification that your number is indeed prime.

Solution by Noether Berlin vs Borussia Munchengoldbach in the Bundesleibniz

The largest prime beginning with the decimal expansion \(2015 \dots\) that we can verify is 201599987. To prove this we use the following primality test:

Pocklington Criterion

Let \(N>1\) be a positive integer. If there exists integers \(a\) and \(q\) such that:
  1. \(q\) is prime, \(q|N-1\) and \(q>\sqrt{N}-1\),
  2. \(a^{N-1} \equiv 1\) mod \(n\),
  3. gcd\( (a^{(N-1)/q}-1,N)=1\)
then \(N\) is prime.

Set \(N = 201599987\). The prime factorisation of \(N-1\) is \(2.7.14399999\) so we have a candidate \[q = 14399999.\] It is prime, divides \(N-1\) and \(\sqrt{N}-1 = 14197.59\dots < 14399999\) so we have a suitable value of \(q\).

We try \(a=2\). Using Maple we see that \(2^{N-1} \equiv 1\) mod \(N\) so the second criterion is satisfied with this value. Now \[\gcd(a^{(N-1)/q}-1,N) = \gcd(16383,N)\] where 16383 has prime factorisation 3.43.127. A trivial check shows that none of 3, 43 or 127 divide \(N\) so 16383 shares no prime factors with \(N\). Hence gcd\((16393,N) = 1\) and condition 3 is satisfied.

So, by the Pocklington Criterion with \(a=2\) and \(q=14399999\), we deduce that 201599987 is prime. (A screenshot of the Maple syntax can be found here)