Number Theory Help: Proving Expression is Equal to Prime Number

AI Thread Summary
The discussion centers on the expression a + bx, where gcd(a, b) = 1, and whether it can equal a prime number for some integer x. Initial attempts to prove this led to confusion about the nature of integers and rational numbers, with participants debating the validity of certain steps in the proof. A reference to Dirichlet's theorem suggests that while proving the existence of primes in such expressions is complex, it is a well-studied topic in number theory. Participants also share resources for further reading, including a PDF on Dirichlet's proof and mention of the sci.math newsgroup for additional insights. The complexity of the problem highlights the challenges faced by beginners in number theory.
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Hi guys, i m just a begineer in number theory.
While solving some questions ,i came across a doubt.

The expression: a+bx
here, gcd(a,b)=1
There always exists a value of x(where x is a integer) such that the above
expression is equal to a prime number.

Can anyone prove the above statement (if it is true).

Also,please suggest some good book on number theory for begineers.
Thanks in advance.
 
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a+bx=p
x(a/x +b)=p
a/x + b=p/x.

P is prime be definiton, so p/x can not be an integer.

a/x + b can not be an integer.
a/x can not be an integer.

Lets let this non integer equal t.
a/x=t
a=xt
Since a has to be an integer, and t is not an integer, x can not be an integer either.

This theorem is false in the natural numbers, or the integers.
 
Thanx for solution,
But I hav a doubt in :
a/x=t
a=xt
Since a has to be an integer, and t is not an integer, x can not be an integer either.

Here ,how can we say that product of a rational number and integer is not a
integer. As,here x is not a prime(assume) may be factor of denominator of t.

Also,a+bx=p
x=(p-a)/b
means;
p=a(mod b)

Here,just studying a special case,
a=1 and b=r(r is a prime)
1+rx=p (r,p are both primes)
rx=p-1
x=(p-1)/r
p=1(mod r)
This equation has integra solutions.
Above is derived from a
Statement that for any prime p there exists a complete residue system modulo n,all whose members are primes.
So, a can be anything from 1,2,3,...r-1
and r,p are primes.
 
Umm I may have gone onto this question intuitively rather than a solid proof.

The product of a rational integer and a non integer can only be another integers if the non integer contains a factor in its denominator. So we have to prove that t is of the form C/nx. But, if it is, the a=c/n. so T =a/x. Since that is true, it is of the form c/nx and that makes my proof wrong >.< but hopefully that leads you in a good direction. Sorry about the mistake
 
Do you have any reason to believe that that is an 'easy' problem? There is a proof by Dirichlet that a series like you have defined contains an infinite number of primes. But the proof uses complex analysis and isn't elementary. One might hope that the job of proving the series contains at least one prime might be easier than showing it contains an infinite number. But I'm really not sure it is.
 
Well Dick,you may be right that my question is not that easy,but as i told that
myself being just a begineer ,i may not have duly reconginsed the depth in the question.
But,my intution says that simple expressions like 1+px(p being a prime)
will definitely yield a prime.But,had no idea that even simple problem like that are not that simple at all.

Well,can you please tell from where can i find that proof by Dirichlet.
 
Here's a pdf I found:

modular.fas.harvard.edu/129/projects/weissman/project.pdf

You might also try to find the alternative proof by Selberg cited in the paper - which doesn't use complex numbers but is "long and not particularly enlightening".
 
Also the newsgroup sci.math is a great place to post questions like this. There are some smart people there.
 
Gib Z said:
a+bx=p
x(a/x +b)=p
a/x + b=p/x.

P is prime be definiton, so p/x can not be an integer.

That's the first mistake in your proof. p/x can be an integer, if x = 1.

Work through an example like 2 + 3.5 = 17 to find more mistakes.
 
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