Hi,
just stuck up with questions on prime numbers.
1) is 5^50 + 5^25 + 1 prime?
2) for how many prime p, p+8 and p+14 also prime?
Thanks
For (2), I would think it hard to prove there are only finitely many such primes. Usually,there is a proof that there are infinitely many such primes. So, I would attempt a proof by contradiction starting with the proposition that there are only finitely many primes $\displaystyle p$ such that $\displaystyle p+8$ and $\displaystyle p+14$ are both prime. Have you attempted this at all?
Such a prime, if it exists, must be of the form $6k + 5$. To see this, note that:
$6k$ is divisible by 6
$6k + 2$ and $6k + 4$ are divisible by 2
$6k + 3$ is divisible by 3.
That leaves only $6k + 1$ and $6k + 5$. If we have $p = 6k+1$, then $p+8 = 6k + 9$ is divisible by 3.
(this is really just saying $p$ is of the form $3n + 2$).
Or, for question 2, with the exception of p=5, try modulus 30 to weed out the 5's later on. XD
All candidates listed: 5, 11, 17, 23, 29.
Clearly 11+14=25, 17+8=25, and 5|25, so the first three can be weeded out.
Now, check p if p=23 or 29 (mod 30)
p=3,p+8=11,p+14=17 has p=3 not congruent to 5 (mod 6). So, that should be amended p=3 or $\displaystyle p \equiv 5\pmod{6}$
Then, you can find congruences mod other primes.
p=5 or
$\displaystyle p\equiv 1,4\pmod{5}$
p=7 or
$\displaystyle p\equiv 1,2,3,4,5\pmod{7}$
p=11 or
$\displaystyle p\equiv 1,2,4,5,6,7,9,10\pmod{11}$
p=13 or
$\displaystyle p\equiv 1,2,3,4,6,7,8,9,10,11\pmod{13}$
For primes $\displaystyle q>13$, either $\displaystyle p=q$ or $\displaystyle p{\not \equiv}q,q-8,q-14\pmod{q}$
Not sure if that helps at all.