Show that the sum of twin primes ## p ## and ## p+2 ## is divisible?

[Math100]

Homework Statement

Show that the sum of twin primes $p$ and $p+2$ is divisible by $12$, provided that $p>3$.

Relevant Equations

None.

Proof:

Suppose $p$ and $p+2$ are twin primes such that $p>3$.
Let $p=2k+1$ for some $k\in\mathbb{N}$.
Then we have $p+(p+2)=2k+1+(2k+1+2)=4k+4=4(k+1)=4m$,
where $m$ is an integer.
Thus, the sum of twin primes $p$ and $p+2$ is divisible by $4$.
Since $p>3$, it follows that $p\equiv 2 \mod 3$ and $p+2\equiv 1 \mod 3$.
This means $p+(p+2)\equiv 0 \mod 3$,
which implies that the sum of twin primes $p$ and $p+2$ is divisible by $3$.
Note that $p+(p+2)$ is divisible by both $3$ and $4$.
Thus $12\mid p+(p+2)$.
Therefore, the sum of twin primes $p$ and $p+2$ is divisible by $12$,
provided that $p>3$.

 

[Mark44]

Homework Statement:: Show that the sum of twin primes $p$ and $p+2$ is divisible by $12$, provided that $p>3$.
Relevant Equations:: None.

Proof:

Suppose $p$ and $p+2$ are twin primes such that $p>3$.
Let $p=2k+1$ for some $k\in\mathbb{N}$.
Then we have $p+(p+2)=2k+1+(2k+1+2)=4k+4=4(k+1)=4m$,
where $m$ is an integer.
Thus, the sum of twin primes $p$ and $p+2$ is divisible by $4$.

So far, so good.

Since $p>3$, it follows that $p\equiv 2 \mod 3$ and $p+2\equiv 1 \mod 3$.

Why does it follow that $p\equiv 2 \mod 3$? Isn't it possible that $p\equiv 1 \mod 3$? Given that p is a prime it obviously can't be true that $p \equiv 0 \mod 3$, but you didn't say anything about the other possibility I pointed out.

This means $p+(p+2)\equiv 0 \mod 3$,
which implies that the sum of twin primes $p$ and $p+2$ is divisible by $3$.
Note that $p+(p+2)$ is divisible by both $3$ and $4$.
Thus $12\mid p+(p+2)$.
Therefore, the sum of twin primes $p$ and $p+2$ is divisible by $12$,
provided that $p>3$.

The last part, assuming you can address my earlier concern, is wordier than necessary.

Instead of this

sum of twin primes $p$ and $p+2$ is divisible by $3$

all you need to say is this: Thus p + p + 2 is divisible by 3.
You've already set things up by positing that p and p+2 are primes, and that p > 3. You don't need to keep repeating this things.

Since you've already established that p + p + 2 is divisible by 4, you can conclude that p + p+2 is divisible by 12.
Once you establish that

 

[Math100]

So far, so good.

Why does it follow that $p\equiv 2 \mod 3$? Isn't it possible that $p\equiv 1 \mod 3$?

Yes, I just realized that both are possible and both should be included, $p\equiv 1 \mod 3$ and $p\equiv 2 \mod 3$. But how should I include both of them in this proof? Should I do/include two cases?

 

[Mark44]

Should I do/include two cases?

Yes. Case 1 is what you already did. Case 2 would be the other possibility that I mentioned.

 

[Mark44]

I've edited my post #2 -- be sure to take another look at it.

 

[Mark44]

BTW, Case 2 is very easy.

 

[Math100]

So I found out that if $p\equiv 1 \mod 3$, then $p+2\equiv 0 \mod 3$. Thus $p+(p+2)\equiv 1 \mod 3$.

 

[Mark44]

So I found out that if $p\equiv 1 \mod 3$, then $p+2\equiv 0 \mod 3$. Thus $p+(p+2)\equiv 1 \mod 3$.

Yes, but what does $p+2\equiv 0 \mod 3$ tell you about p + 2?

 

[Math100]

Yes, but what does $p+2\equiv 0 \mod 3$ tell you about p + 2?

That the prime $p+2$ is divisible by 3.

 

[Mark44]

Like I said, Case 2 is pretty easy.

 

[Math100]

So I just have to mention/include that if $p\equiv 1 \mod 3$, then $p+2\equiv 0 \mod 3$ for case #2?

 

[gbosdet]

So far, so good.

Why does it follow that $p\equiv 2 \mod 3$? Isn't it possible that $p\equiv 1 \mod 3$? Given that p is a prime it obviously can't be true that $p \equiv 0 \mod 3$, but you didn't say anything about the other possibility I pointed out.

If $p\equiv 1 \mod 3$ then $p+2\equiv 0 \mod 3$ and p+2 isn't prime, violating that assumption

 

[Math100]

If $p\equiv 1 \mod 3$ then $p+2\equiv 0 \mod 3$ and p+2 isn't prime, violating that assumption

Now I see.

 

[Mark44]

Another way to look at what we're calling Case 1 is that if $p \equiv 2 \mod 3$, that's the same as saying $p \equiv -1 \mod 3$.
So $p + 2 \equiv +1 \mod 3$, and $p + p+2 \equiv 0 \mod 3$.

I'm not saying that either way has an advantage over the other.

 

[Math100]

Suppose $p$ and $p+2$ are twin primes such that $p>3$.
Let $p=2k+1$ for some $k\in\mathbb{N}$.
Then we have $p+(p+2)=2k+1+(2k+1+2)=4k+4=4(k+1)=4m$,
where $m$ is an integer.
Thus $4\mid p+(p+2)$.
Since $p>3$, it follows that $p\equiv 1 \mod 3$ or $p\equiv 2 \mod 3$.
Now we consider two cases.
Case #1: Suppose $p\equiv 1 \mod 3$.
Then we have $p+2\equiv 0 \mod 3$.
This is a contradiction because $3\nmid p+2$,
given the fact that $p+2$ is twin prime.
Case #2: Suppose $p\equiv 2 \mod 3$.
Then we have $p+2\equiv 1 \mod 3$.
Thus $p+(p+2)\equiv 0 \mod 3$,
which implies that $3\mid p+(p+2)$.
Note that $p+(p+2)$ is divisible by both $3$ and $4$.
Thus $12\mid p+(p+2)$.
Therefore, the sum of twin primes $p$ and $p+2$ is divisible by $12$,
provided that $p>3$.

 

[Mark44]

Looks good, overall. If I were doing this problem, I would add the parts in green, and delete the parts that are struck out.

Suppose $p$ and $p+2$ are twin primes such that $p>3$.
Let $p=2k+1$ and $p + 2 = 2k + 3$ for some $k\in\mathbb{N}$.
Then we have ## p+(p+2)=2k+1+(2k+1+2)=4k+4=4(k+1)=4m $, where$ m ## is an integer.
Thus $4\mid p+(p+2)$.
Since $p>3$, it follows that Now either $p\equiv 1 \mod 3$ or $p\equiv 2 \mod 3$.
Now we consider two cases.
Case #1: Suppose $p\equiv 1 \mod 3$.
Then we have $p+2\equiv 0 \mod 3$.
This is a contradiction because $3\nmid p+2$,
given the fact that $p+2$ is twin prime.
Case #2: Suppose $p\equiv 2 \mod 3$.
Then we have $p+2\equiv 1 \mod 3$.
Thus $p+(p+2)\equiv 0 \mod 3$,
which implies that so $3\mid p+(p+2)$.
Note that $p+(p+2)$ is divisible by both $3$ and $4$.
Thus $12\mid p+(p+2)$.
Therefore, the sum of twin primes $p$ and $p+2$ is divisible by $12$,
provided that $p>3$.

An elegant proof is one that says everything that needs to be said, but without an extraneous wordage. Several people who have made comments on your other threads have been steering you in this direction.

 

[Orodruin]

Why does it follow that p≡2mod3? Isn't it possible that p≡1mod3?

No, if $p\equiv 1 \mod 3$ then $p+2 \equiv 0 \mod 3$ so $p+2$ is not prime. It therefore does follow that $p\equiv 2\mod 3$. This follows directly from the basic assumptions of the problem and I don’t think it qualifies to be called its own case. I would just say:
Since p > 3, p=2 mod 3 as p=0 mod 3 would not be prime and p=1 mod 3 would imply that p+2=0 mod 3 would not be prime.

 

[Mark44]

Why does it follow that p≡2mod3? Isn't it possible that p≡1mod3?

No, if $p\equiv 1 \mod 3$ then $p+2 \equiv 0 \mod 3$ so $p+2$ is not prime. It therefore does follow that $p\equiv 2\mod 3$. This follows directly from the basic assumptions of the problem and I don’t think it qualifies to be called its own case. I would just say:
Since p > 3, p=2 mod 3 as p=0 mod 3 would not be prime and p=1 mod 3 would imply that p+2=0 mod 3 would not be prime.

Yes, I understand all of the above. My post was to get the OP to understand that he had not addressed (and then discounted) the possibility that $p \equiv 1 \mod 3$ might be true. As has already been shown in this thread, p and p + 2 can't be twin primes if $p \equiv 1 \mod 3$.