Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a system of intervals: $[A_1, +\infty)$; $[A_2, +\infty)$; $[A_3, +\infty)$; ... $A_1 < A_2 < A_3 < ...$ etc.

Let $NP(i)$ denote the number of primes of the form "s^2+d" , "d" is constant integer, (+infinity is also acceptable) occurring in interval $[A_i, +\infty)$, $i = 1, 2, 3,$ ... Let $NP(1) + NP(2) + NP(3) + ... = + \infty$.

Is deduction that there has to exist such number $k$ that $ NP(k) = +\infty$ OK ?.

Consider that intersection of all intervals $[A_i, +\infty)$, $i = 1, 2, 3,$ ... is null set.

share|improve this question
1  
What does NP(k) mean here, and what does it have to do with your system of intervals? –  Henning Makholm Apr 4 '13 at 16:50
    
What does $<$ means for the intervals? Is it $[$ ? –  1015 Apr 4 '13 at 16:54
    
NP(i) denote the number of primes (infinite is also acceptable) occurring in i-th interval <Ai,+∞), i=1,2,3, ... –  pgalik Apr 4 '13 at 16:59
    
mark "<" means that interval is closed –  pgalik Apr 4 '13 at 17:01
    
I have to note that primes in interval <Ai,+∞) have the form s^2+d, where "d" is some constant integer and "s" is integer variable –  pgalik Apr 4 '13 at 17:06

1 Answer 1

Yes, this inference is correct.

If $NP(1)<+\infty$, then only a finite number of primes of the form $s^2+d$ occur in $[A_1,+\infty)$. Suppose that there are $r$ such primes and that the biggest such prime is $P$. Then since $A_1<A_2<\cdots$, there must be some $k$ with $A_k>P$. Now $$ r=NP(1)\ge NP(2)\ge NP(3)\ge\cdots \ \ \text{and}\ \ NP(k)=NP(k+1)=\cdots=0, $$ so $$ NP(1)+NP(2)+NP(3)+\cdots\le r(k-1) <+\infty, $$ a contradiction. Therefore, under the assumptions given, $NP(1)+NP(2)+\cdots=+\infty$ implies that $NP(1)=+\infty$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.