# Thread: Does There Exists Three Consecutive Odd Positive Integers That Are Primes

1. ## Does There Exist Three Consecutive Odd Positive Integers That Are Primes

Hello,

The exact question is, Prove or disprove that there are three consecutive odd positive integers that are primes, that is, odd primes of the form $p$, $p+2$, and $p+4$.

At first glance, I did not realize that this statement was an existentially quantified one, that one example would prove this true--the example being 3, 5, and 7. So, I did a full-fledged proof and am wondering if I did it correctly, notwithstanding its necessity.

Let $p$ be an odd integer that is prime, then $p = 2k + 1$, where $k \in \mathcall{Z}$.

$p+2=(2k + 1)+2 \rightarrow p+2= 2k + 3$ this is prime, because the only factors are $2k+3$ and 1.

$p+4 = (2k+1) + 4 \rightarrow p+4 = 2k + 5$ this is prime, because the only factors are $2k+5$ and 1.

As I type this, I realize there is something wrong with it: there seems to be a problem with letting $k$ be any integer. Also, if this proof is correct, does this prove that there exists three consecutive odd positive integers that are primes, but doesn't give any way of determining them? If this is absolutely wrong, is there some way to proof this, other than giving an example?

2. ## Re: Does There Exist Three Consecutive Odd Positive Integers That Are Primes

Originally Posted by Bashyboy
Hello,

The exact question is, Prove or disprove that there are three consecutive odd positive integers that are primes, that is, odd primes of the form $p$, $p+2$, and $p+4$.

At first glance, I did not realize that this statement was an existentially quantified one, that one example would prove this true--the example being 3, 5, and 7. So, I did a full-fledged proof and am wondering if I did it correctly, notwithstanding its necessity.

Let $p$ be an odd integer that is prime, then $p = 2k + 1$, where $k \in \mathcall{Z}$.
p=
$p+2=(2k + 1)+2 \rightarrow p+2= 2k + 3$ this is prime, because the only factors are $2k+3$ and 1.
No, that last statement is not true. That would be true only is 2k+ 3 were prime, which is what you are trying to show. For example, if k happened to be 6, p= 2k+1= 13, a prime number. But p+ 2= 2k+ 3= 15.

$p+4 = (2k+1) + 4 \rightarrow p+4 = 2k + 5$ this is prime, because the only factors are $2k+5$ and 1.

As I type this, I realize there is something wrong with it: there seems to be a problem with letting $k$ be any integer. Also, if this proof is correct, does this prove that there exists three consecutive odd positive integers that are primes, but doesn't give any way of determining them? If this is absolutely wrong, is there some way to proof this, other than giving an example?

3. ## Re: Does There Exist Three Consecutive Odd Positive Integers That Are Primes

Originally Posted by Bashyboy
The exact question is, Prove or disprove that there are three consecutive odd positive integers that are primes, that is, odd primes of the form $p$, $p+2$, and $p+4$.

At first glance, I did not realize that this statement was an existentially quantified one
Well, yes: a statement that starts with "There is (are)" or "There exist(s)" is existentially quantified.

Originally Posted by Bashyboy
that one example would prove this true--the example being 3, 5, and 7. So, I did a full-fledged proof and am wondering if I did it correctly, notwithstanding its necessity.
Yes, a single example is all one needs. Strictly speaking, in proving ∃x P(x), one has to not just exhibit a witness x but also prove P(x). In this case, you need to exhibit three numbers and then prove that they are positive, odd and prime. Often, as in this case, proving P(x) is trivial, but it's not always.

Originally Posted by Bashyboy
Let $p$ be an odd integer that is prime, then $p = 2k + 1$, where $k \in \mathcall{Z}$.

$p+2=(2k + 1)+2 \rightarrow p+2= 2k + 3$ this is prime, because the only factors are $2k+3$ and 1.
How do you know this claim about the factors? What if k = 6?

Originally Posted by Bashyboy
As I type this, I realize there is something wrong with it: there seems to be a problem with letting $k$ be any integer.
You are right in feeling this way: you can't prove this for all k.

Originally Posted by Bashyboy
Also, if this proof is correct, does this prove that there exists three consecutive odd positive integers that are primes, but doesn't give any way of determining them? If this is absolutely wrong, is there some way to proof this, other than giving an example?
There are two ways to prove ∃x P(x): exhibiting a concrete x and proving P(x) (this is called a constructive proof) or a proof by contradiction, i.e., showing that ∀x ¬P(x) is false. For simple properties P(x) (those that don't contain quantifiers, which is the case for this problem) there is always a way to convert a proof by contradiction into a constructive proof, though sometimes it may be unclear how a proof by contradiction contains a witness. Thus, every proof of this particular claim basically works by giving concrete three numbers. An essentially non-constructive proof by contradiction is possible only for complex P or when additional axioms are used.

4. ## Re: Does There Exists Three Consecutive Odd Positive Integers That Are Primes

So, for this particular problem, it's really only possible to give concrete examples to prove the statement true?

5. ## Re: Does There Exists Three Consecutive Odd Positive Integers That Are Primes

Originally Posted by Bashyboy
So, for this particular problem, it's really only possible to give concrete examples to prove the statement true?
Yes.

6. ## Re: Does There Exists Three Consecutive Odd Positive Integers That Are Primes

Thank you both very much, emakarov and HallsofIvy.

7. ## Re: Does There Exists Three Consecutive Odd Positive Integers That Are Primes

It's worth noting that {3,5,7} is the only occurrence of three consecutive primes. Any three consecutive arithmetic terms (of positive integers) will have a term that is divisible by three. The only way to make all three of them prime would be to have the only prime multiple of three (3) be included.