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

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- Jul 31st 2014, 11:15 AMnikhilprime numbers
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 - Jul 31st 2014, 05:49 PMSlipEternalRe: prime numbers
1) $\displaystyle x^{50}+x^{25}+1$ is divisible by $\displaystyle x^2+x+1$ in $\displaystyle \Bbb{Z}[x]$, so no, it is not prime. $\displaystyle 5^2+5+1=31$ making that number divisible by 31.

2) haven't thought about it - Aug 1st 2014, 11:55 PMnikhilRe: prime numbers
could you explain how you figured x^2+x+1

will divide the expression x^50+x^25+1 - Aug 2nd 2014, 06:06 AMSlipEternalRe: prime numbers
It is an expression of the form $\displaystyle x^{2k}+x^{k}+1$, which is divisible by $\displaystyle x^2+x+1$ for all $\displaystyle k$.

- Aug 2nd 2014, 06:39 AMIdeaRe: prime numbers
- Aug 2nd 2014, 08:27 AMDevenoRe: prime numbers
- Aug 2nd 2014, 08:34 AMSlipEternalRe: prime numbers
Oh, I remembered proving something about it in my algebra class years ago. That must have been it. My memory is not what it used to be.

- Aug 2nd 2014, 09:30 AMDevenoRe: prime numbers
- Aug 2nd 2014, 09:43 AMIdeaRe: prime numbers
- Aug 2nd 2014, 11:52 AMDevenoRe: prime numbers
- Aug 2nd 2014, 12:33 PMSlipEternalRe: prime numbers
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?

- Aug 3rd 2014, 05:40 PMjohngRe: prime numbers
I was puzzled by the divisibility fact mentioned above. Since others might also have been puzzled, here's a proof:

http://i57.tinypic.com/15h036h.png - Aug 3rd 2014, 09:36 PMDevenoRe: prime numbers
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 $3n + 2$). - Aug 4th 2014, 03:56 PMdennydenglerRe: prime numbers
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) - Aug 6th 2014, 07:15 AMSlipEternalRe: prime numbers
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.