Can the sum of two primes be prime if both primes are odd?

  • Thread starter Thread starter cragar
  • Start date Start date
  • Tags Tags
    Primes Proof
cragar
Messages
2,546
Reaction score
3

Homework Statement



If the sum of two primes is prime, then one of the primes must be 2.

The Attempt at a Solution


Proof:
Since all primes bigger than 2 are odd the only way to get a sum of two primes to be odd is to add an odd prime with an even prime.
Let y be an odd prime such that there exists and integer q so that y=2q+1, and then we will add this to 2 giving us a new number k such that k=2+(2q+1)=2q+3 which is not divisible by 2 therefore it is odd. Suppose for the sake of contradiction that both of the primes were odd and when added together were prime.
Let integers T and P be given that are odd primes. And T=2s+1 , where s is an integer. And P=2d+1. Now if we add T+P , we get that T+P= (2s+1)+(2d+1)=2s+2d+2=2(s+d+1) , which is divisible by 2 and is not prime by definition and is a contradiction.
My proof is kinda choppy and i kinda used 2 methods in the proof. Which would be better, to do a proof by contradiction of a direct proof?
 
Physics news on Phys.org
I'm a little confused. You start by quoting a perfectly good proof idea:

Since all primes bigger than 2 are odd the only way to get a sum of two primes to be odd is to add an odd prime with an even prime.

(in fact, I would have accepted this sentence as a proof! But maybe you're still at the stage where you need to flesh out the details)

but then the rest of your proof doesn't seem to have anything to do with the proof idea.
 
ya the second part doesn't really relate to the first part. But you said the first part is fine. So ill just stick with it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top