# UKMT Special (Problem $1$)

Show that there are at least $3$ primes that are less than $200$ that satisfy the system of equations:

$p + 2$

$p + 6$

$p + 8$

$p + 12$

and result in primes.

Also show that there is only one one prime that satisfy the system of equations:

$q + 2$

$q + 6$

$q + 8$

$q + 12$

$q + 14$

and result in primes.

[UKMT BMO $2019$, Q$1$]

Note by Yajat Shamji
2 weeks, 6 days ago

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- 2 weeks, 6 days ago

You have until next Saturday, $3:00$pm!

- 2 weeks, 6 days ago

I'm interpreting the question as "Find all primes $p$ such that the list of expressions above are all prime". However, there are no such primes that even satisfy the first set of requirements! For $p \ge 2$, we notice that from Pigeonhole Principle, there must be a number from the first $3$ statements that is divisible by $3$. One can easily check that the number is composite.

The only number that actually satisfies this restriction is $p = 1$, and even still $1$ is not a prime.

Since the weaker condition cannot be satisfied, neither can the stronger condition involving $q$.

- 2 weeks, 5 days ago

There is a solution!

I solved it myself!

- 2 weeks, 4 days ago

Oops!

I accidentally wrote the wrong system of equations!

Re-solve it, @Elijah L

- 2 weeks, 4 days ago

Note that from the first set of equations, $p$ must be $0$, $3$, or $4 \pmod 7$ for sufficiently large $p$. Looking at it $\pmod 5$, the only permissible residues mod $5$ are $0$ and $1$. From the same logic, the number must also be $2 \pmod 3$. Checking small cases leaves that $p = 5$ is a solution.

Finding primes that are $3$ or $4 \pmod 7$ and congruent to $1 \pmod 5$ is all that is left for part 1. We can strengthen this condition and say that the last digit of the prime must be $1$, as if it is $6$ the number is not prime. The next prime is $p = 11$, and the next is $p = 101$. Hence, there are a minimum of $3$ primes that satisfy the first set of requirements.

For the part of the question involving $q$, note that $5$ is the only number that satisfies the requirements. This is because the only permissible residue mod $5$ is 0, and $5$ is the only prime that is congruent to $0 \pmod 5$.

- 2 weeks, 4 days ago

Correct, but different method.

- 2 weeks, 3 days ago