Hello,
I need a few solutions here guys,

1. Let n> 1 be an integer. Prove that if 2ⁿ +n² is a prime, then nº 3(mod 6)

2. Prove that for any prime p there exists an infinite number of positive integers m such that p is a divisor of 2^m+1 +3^m -17

3. evaluate
1/1 + 1/2³ + 1/3³ + ....

love arun

1. Let y = 2ⁿ + n² . Then n cannot be even, for if it is, then y would be divisible by 2. Therefore n = 6k ± 1 or 6k + 3. Put n = 6k + 1 into your equation, and you get

y = 2^{6k + 1} + (6k + 1)² = 2^{6k + 1} + 36k² + 12k + 1 = 2(2^6k + 18k² + 6k) + 1.

Now, all primes greater than 3 are of the form 6k ± 1 (can you see why?). Therefore for y to be prime, 2^6k + 18k² + 6k must be a multiple of 3. Since 18k² + 6k is a multiple of 3, then so must 2^6k be, but this is clearly not true - contradiction. The other 2 cases follow in the same way.

Regards,
Olof

Ah, just noticed that you also need to consider the case y = 2(2^6k + 18k² + 6k) + 1 = 6m - 1.

Regards,
Olof

$\zeta (2)={\pi }^2/6$ and

$\zeta (4)={\pi }^4/90$.

Olof

Olof is correct that no closed form is know for $\zeta (3)$. It isknown that $\zeta (3)$ is irrational (this was proved by Apéry) but not much more is known.

Nice method Michael!

Olof

Thanks, but I should also have said that the cases p = 2 and p = 3 need to be treated separately (they are very easy though).

Olof,

I understood that n cannot be even,but then n must be all odds.
Shouldn't n = 2k+1 ?? (since n> 1)

please explain this to me.

love arun

Michael,

i will just now be trying your method.
i will tell you tomorrow if i don't get it.
is it ok??

love arun

Also,
i would like to have some info on riemann zeta function.

love arun

Well, 2k + 1 is one way of writing an odd number, but all odd numbers can also be represented by 6k + 1, 6k + 3 or 6k + 5 (which is equivalent to 6k - 1). Since we have to look at things mod 6, I thought it would make sense to consider these cases from the start, to make it a bit simpler.

Regards,
Olof

P.S.
I've been very sloppy in my use of k's and m's etc. in my post above, sorry for that. I'm sure you'll be able to get the general idea though.

Olof,
I see where you are getting at.

thanks
love arun

Michael my man,
that was an absolute gorgeous answer.

I wonder how did you come up with
m = 2+ l (p-1)

love arun

Sorry; didn't see your other post. Fermat's little theorem is the key. If you define:

f(n) = 2^m+1 + 3^m = 2*2^m + 3^m

Then you have:

f(n + p-1) = 2*2^{n + p-1} + 3^{m + p-1}

So:

f(n + p-1) = 2*2ⁿ + 3^m (mod p)

since 2^p-1 = 3^p-1 = 1 (mod p) by Fermat's Little Theorem.

So f(n + p-1) = f(n) (mod p).

Therefore f(n) is periodic (mod p) with period p-1.

We are trying to show that the equation f(n) = 17 (mod p) has infinitely many solutions. But as f(n) is periodic we only have to find one solution for n. It will then immediately follow there are infinitely many. So let's search for a solution.

f(0) = 3
f(1) = 7
f(2) = 17

So f(2) = 17 (mod p) and therefore f(2 + lambda(p-1)) = 17 (mod p). This is the required result.