As you posted it - the term in question is:3k+1 is prime number
Prove that for all positive integers n there is no such k which would make always a prime number in this 2^n*3k+1
I thought about that too. But that question is just plain idiotic.I suspect that the expression should be 2^n*(3k+1).
Clearly 2 | 2^n*(3k+1), so 2^n*(3k+1) is not prime. Is that what you want?
Lupik - you came back to visit the site/question ~8 hours ago - yet, you did not respond !!3k+1 is prime number
Prove that for all positive integers n there is no such k which would make always a prime number in this 2^n*3k+1
ItYour translation is very poor. Do you mean you are supposed to prove
[MATH]\text {Given that } k,\ n \in \mathbb Z^+ \text { and } 3k + 1 \text { is prime} \implies\\ (2^n * 3k) + 1 \text { is not prime.}[/MATH]But that simply is not true.
If k = 2, then 3k + 1 = 7, which is prime.
If n = 1, then 2^1 * 3k + 1 = 2 * 6 + 1 = 13, which is prime.
The goal is to prove that you cant find any k for which whould ve this expression always prime number for all value of positive integer n.Your translation is very poor. Do you mean you are supposed to prove
[MATH]\text {Given that } k,\ n \in \mathbb Z^+ \text { and } 3k + 1 \text { is prime} \implies\\ (2^n * 3k) + 1 \text { is not prime.}[/MATH]But that simply is not true.
If k = 2, then 3k + 1 = 7, which is prime.
If n = 1, then 2^1 * 3k + 1 = 2 * 6 + 1 = 13, which is prime.
Ahh i get it it now. Prove thatThe expression is
[MATH]x=(2^n*3k)+1[/MATH]n is starting from zero and going to infinity and you have to show that you cant find positive integer k for which would all values of expression
n=1
[MATH](2^1*3k)+1[/MATH]n=2
[MATH](2^2*3k)+1[/MATH]...
n=infinity
make always a prime number
Yeah you got it, I was thinking about this a lot and still I dont know how to prove that.Ahh i get it it now. Prove that
[MATH]\text {There is no positive integer k such that}\\ (2^n * 3k) + 1 \text { is prime for every positive integer } n.[/MATH]Consequently, my example was flawed.
[MATH]k = 2 \text { and } n = 2 \implies (2^n * 3k) + 1 = (4 * 6) + 1 = 25, \text { not prime.}[/MATH]
Do I finally have it?
Also there is still conditionI have not found a proof either.
But if k = 1 and n = 1, x = 7
k = 1 and n = 2, x = 13
k = 1 and n = 3, x = 25 perfect square
k = 2 and n = 1, x = 13
k = 2 and n= 2, x = 25 perfect square
k = 2 and n = 3, x = 49 perfect square
k = 3 and n = 1, x = 19
k = 3 and n = 2, x = 37
k = 3 and n = 3, x = 73
k = 3 and n = 4, x = 137
k = 3 and n = 5, x = 289 perfect square
Those perfect squares look intriguing. (Of course, it may be a false clue.)
No, k does not have to be even because, as you have framed the question, we are dealing withAlso there is still condition
[MATH]3k+1=[/MATH]Prime number
So k must be even number.
Nice, great work!In fact we don't need [MATH]3k+1[/MATH] to be prime.
Prove that [MATH]\,\forall k\in \mathbb{Z^+}\hspace1ex \exists n: \, k(2^n)+1[/MATH] is composite
Obviously true for k=1 (e.g. n=3)
Consider k>1, then (a) [MATH]\exists[/MATH] p prime >2: [MATH]\,p|(k+1)[/MATH] or (b) [MATH]k+1=2^m[/MATH] for some [MATH]m>1[/MATH]
Taking the first case, (a):
[MATH](k+1)2^n \equiv0[/MATH] mod p
[MATH]k(2^n)+1 + 2^n -1 \equiv 0[/MATH] mod p
Let [MATH]n=p-1[/MATH][MATH]k(2^{p-1}) + 1 + 2^{p-1} -1 \equiv 0[/MATH] mod p
[MATH]2^{p-1}-1 \equiv 0[/MATH] mod p (since p>2)
[MATH]\therefore k(2^{p-1}) + 1 \equiv 0[/MATH] mod p
[MATH]k(2^{p-1}) + 1>2^p>p \hspace1ex \therefore k(2^{p-1})+1[/MATH] is composite.
Taking the second case, (b):
[MATH]k+1=2^m[/MATH] (m>1 since k>1) and we want to find n: [MATH]k(2^n)+1[/MATH] is composite, i.e. [MATH](2^m-1)(2^n)+1[/MATH] is composite.
[MATH](2^m-1)(2^n)+1 = 2^{m+n}-2^n+1[/MATH]
Let n=m+2
[MATH]2^{m+n}-2^n+1=2^{2(m+1)}-2(2^{m+1})+1\\ =(2^{m+1}-1)^2[/MATH]which is composite (m>1).