1. ## Prime numbers

Try this ... from AIMO 2007 Intermediate Paper

Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.

I don't even know where to start!

2. Originally Posted by BG5965
Try this ... from AIMO 2007 Intermediate Paper

Find a prime, p, with the property that for some larger prime number, q, both 2q - p and 2q + p are prime numbers. Prove that there is only one such prime p.

I don't even know where to start!
i'll take p=3..

for q=5: we have 7 and 13
for q=7: 11 and 17
for q=13: 23 and 29
for q=17: 31 and 37
....

maybe, i'll try to come up with the proof on uniqueness soon...

3. Any prime p > 3 is of the form $6k\pm1$ for some integer $k\ge1$. For p > 3, you can try all combinations of $p=6k_1\pm1$ and $q=6k_2\pm1$ and verify that $2q\pm p$ is never prime.

If p = 2, then 2q+p and 2q-p are both even and neither is equal to p, so neither is prime.

Hence p = 3 is the only prime satisfying the given property.

4. wow!

but how to show that any prime p>3 is of the form $6k \pm 1$ for some integer $k\geq1$?

5. Originally Posted by kalagota
wow!

but how to show that any prime p>3 is of the form $6k \pm 1$ for some integer $k\geq1$?
All odd integers are of the form $6k\pm 1$ or $6k \pm 3$, the latter are all divisible by $3$ and so if greater than $3$ itself are not prime.

RonL

6. Originally Posted by kalagota
but how to show that any prime p>3 is of the form $6k \pm 1$ for some integer $k\geq1$?
If a prime is greater than 3, then it's not divisible by 2 or 3. Therefore it cannot be of the form 6k, 6k+2, 6k+3 or 6k-2, otherwise it would be divisible by 2 or 3 (or both). Hence all primes greater than 3 must be of the form 6k+1 or 6k-1.