# 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)