For n>3, show that the integers n, n+2, n+4 cannot all be prime

[Math100]

Homework Statement

For $n>3$, show that the integers $n, n+2, n+4$ cannot all be prime.

Relevant Equations

None.

Proof:

Let $n>3$ be an integer.
Applying the Division Algorithm produces:
$n=3q+r$ for $0\leq r< 3$,
where there exist unique integers $q$ and $r$.
Suppose $n$ is prime.
Then $n=3q+1$ or $n=3q+2$, because $n\neq 3q$.
Now we consider two cases.
Case #1: Suppose $n=3q+1$ for some $q\in\mathbb{N}$.
That is, $n\equiv 1 \mod 3$.
Then $n+2\equiv 0 \mod 3$ and $n+4\equiv 2 \mod 3$.
Thus, the integer $n+2$ cannot be prime.
Case #2: Suppose $n=3q+2$ for some $q\in\mathbb{N}$.
That is, $n\equiv 2 \mod 3$.
Then $n+2\equiv 1 \mod 3$ and $n+4\equiv 0 \mod 3$.
Thus, the integer $n+4$ cannot be prime.
Therefore, for $n>3$, the integers $n, n+2, n+4$ cannot all be prime.

 

[jbriggs444]

Homework Statement:: For $n>3$, show that the integers $n, n+2, n+4$ cannot all be prime.
Relevant Equations:: None.

Proof:

Let $n>3$ be an integer.
Applying the Division Algorithm produces:
$n=3q+r$ for $0\leq r< 3$,
where there exist unique integers $q$ and $r$.
Suppose $n$ is prime.
Then $n=3q+1$ or $n=3q+2$, because $n\neq 3q$.
Now we consider two cases.
Case #1: Suppose $n=3q+1$ for some $q\in\mathbb{N}$.
That is, $n\equiv 1 \mod 3$.
Then $n+2\equiv 0 \mod 3$ and $n+4\equiv 2 \mod 3$.
Thus, the integer $n+2$ cannot be prime.
Case #2: Suppose $n=3q+2$ for some $q\in\mathbb{N}$.
That is, $n\equiv 2 \mod 3$.
Then $n+2\equiv 1 \mod 3$ and $n+4\equiv 0 \mod 3$.
Thus, the integer $n+4$ cannot be prime.
Therefore, for $n>3$, the integers $n, n+2, n+4$ cannot all be prime.

Did you use the given that $n > 3$ anywhere? If not, that hints at a bit of missing reasoning.

 

[PeroK]

Homework Statement:: For $n>3$, show that the integers $n, n+2, n+4$ cannot all be prime.

In one sentence explain why that statement is true.

 

[fresh_42]

This was good advice in post #3. What was your plan?

For e.g. I distinguish between odd and even numbers $n$ and used the fact that $2$ and $3$ are coprime. You used it, too, but hidden in your argument. However, you did not use that $n$ is prime, so why mention it? And where did you use $n>3$?

 

[Math100]

This was good advice in post #3. What was your plan?

For e.g. I distinguish between odd and even numbers $n$ and used the fact that $2$ and $3$ are coprime. You used it, too, but hidden in your argument. However, you did not use that $n$ is prime, so why mention it? And where did you use $n>3$?

How would you distinguish between odd and even numbers $n$ and used the fact that $2$ and $3$ are coprime? By expressing $n=2k$ for even and $n=2k+1$ for odd?

 

[Math100]

Did you use the given that $n > 3$ anywhere? If not, that hints at a bit of missing reasoning.

By mentioning $q\in\mathbb{N}$.

 

[jbriggs444]

By mentioning $q\in\mathbb{N}$.

Not good enough. $1 \in \mathbb{N}$

 

[fresh_42]

How would you distinguish between odd and even numbers $n$ and used the fact that $2$ and $3$ are coprime? By expressing $n=2k$ for even and $n=2k+1$ for odd?

Yes. The case $n=2k$ is trivial since all three numbers are even in that case. Remains the case $n=2k+1$. Then the numbers in question are $2k+1\, , \,2k+3\, , \,2k+5.$

How can you see (without considering each case!) that at least one of them is divisible by three?

 

[Math100]

Yes. The case $n=2k$ is trivial since all three numbers are even in that case. Remains the case $n=2k+1$. Then the numbers in question are $2k+1\, , \,2k+3\, , \,2k+5.$

How can you see (without considering each case!) that at least one of them is divisible by three?

$3\mid 2k+1$

 

[fresh_42]

$3\mid 2k+1$

No.

We can only conclude that three divides one of these numbers. We cannot know which one. What are the remainders modulo three?

 

[Math100]

No.

We can only conclude that three divides one of these numbers. We cannot know which one. What are the remainders modulo three?

$n\equiv 1 \mod 3$ and $n\equiv 2 \mod 3$.

 

[fresh_42]

$n\equiv 1 \mod 3$ and $n\equiv 2 \mod 3$.

How that? We have the three numbers $2k+1\, , \,2k+3\, , \,2k+5.$ Assume that $2k\equiv r \mod 3.$ Then we have $2k+1\equiv r+1\mod 3\, , \,2k+3\equiv r\mod 3\, , \,2k+5\equiv r+2\mod 3.$ In the set $\{r+1\, , \,r\, , \,r+2\}$ is always exactly one number that is divisible by three, regardless what $r$ actually is.

 

[Math100]

Suppose $n>3$ is an integer.
Now we consider two cases.
Case #1: Let $n$ be an even integer.
That is, $n=2k$ for $k>1$.
Then we have $n+2=2k+2$ and $n+4=2k+4$.
Note that all integers $n, n+2, n+4$ are even.
Thus, all integers $n, n+2, n+4$ are composites.
Case #2: Let $n$ be an odd integer.
That is, $n=2k+1$ for $k>1$.
Then we have $n+2=2k+3$ and $n+4=2k+5$.
Suppose that $2k\equiv r \mod 3$.
Then we have $2k+1\equiv r+1 \mod 3$, $2k+3\equiv r \mod 3$ and
$2k+5\equiv r+2 \mod 3$.
Thus, there exists exactly one integer that is divisible by $3$.
Therefore, for $n>3$, the integers $n, n+2, n+4$ cannot all be prime.

 

[jbriggs444]

@fresh_42 is suggesting that you are making yourself work harder than you need to. There is no need to consider extra cases.

Note that you have also not used the assumption that $n > 3$ which means that you have overlooked two crucial loopholes in the "proof".

 

[WWGD]

Just note there is one multiple of 3 that not prime. Which one? Maybe you meant to say this but we didn't notice.

 

[PeroK]

Suppose $n>3$ is an integer.
Now we consider two cases.
Case #1: Let $n$ be an even integer.

Then $n$ is not prime. There is no need to go further.

That is, $n=2k$ for $k>1$.
Then we have $n+2=2k+2$ and $n+4=2k+4$.
Note that all integers $n, n+2, n+4$ are even.
Thus, all integers $n, n+2, n+4$ are composites.

The above is unnecessary.

Case #2: Let $n$ be an odd integer.
That is, $n=2k+1$ for $k>1$.

That really should be $k > 2$, as $n > 3$.

Then we have $n+2=2k+3$ and $n+4=2k+5$.
Suppose that $2k\equiv r \mod 3$.
Then we have $2k+1\equiv r+1 \mod 3$, $2k+3\equiv r \mod 3$ and
$2k+5\equiv r+2 \mod 3$.
Thus, there exists exactly one integer that is divisible by $3$.

I can't see the relationship between your calculations and your conclusion. It looks somewhat random to me.

Therefore, for $n>3$, the integers $n, n+2, n+4$ cannot all be prime.

True, but where did you prove it?

 

[jbriggs444]

Then $n$ is not prime. There is no need to go further.

Be careful. The fact that $n$ is even does not, by itself, preclude the possibility that $n$ is prime.

 

[PeroK]

Be careful. The fact that $n$ is even does not, by itself, preclude the possibility that $n$ is prime.

That's not the only condition on $n$ in the quoted text. There is also $n > 3$.

 

[fresh_42]

That's not the only condition on $n$ in the quoted text. There is also $n > 3$.

Or alternatively: If $n=2$ then $n+2\neq 2$ and even, hence not prime.

Nevertheless, @jbriggs444 was right:

The fact that $n$ is even does not, by itself,

 

[pbuk]

Why are we discussing odd and even cases?

1. Exactly one of $\{ n, n+1, n+2 \}$ must be divisible by $3$.
2. $n + 1 = 3k \implies n + 4 = 3(k+1)$ so exactly one of $\{ n, n+4, n+2 \}$ must be divisible by $3$.
3. The only prime that is divisible by $3$ is $3$ itself so for $n > 3$ at least one of $\{ n, n+2, n+4 \}$ must be composite.

 

[PeroK]

Or alternatively: If $n=2$ then $n+2\neq 2$ and even, hence not prime.

It's stated in the problem that $n >3$.

 

[fresh_42]

It's stated in the problem that $n >3$.

Yes. Again, @jbriggs444 was right. It is not sufficient by itself. Why do you even debate?

 

[PeroK]

Yes. Again, @jbriggs444 was right. It is not sufficient by itself. Why do you even debate?

Because I'm right.

 

[jbriggs444]

It is often difficult to decide how much to explicitly state and how much to leave out. If one explicitly states too much, it can be insulting to ones audience. If one leaves out too much, the audience may insult the speaker.

So I apologize to @PeroK for doubting the presence of the step not explicitly shown.

 

[fresh_42]

Because I'm right.

I think you confuse right and stubborn.

The case of even $n$ requires one additional argument. Several are possible, but $n\equiv 0 \mod 2$ alone is not sufficient. The additional argument, $n>3$ or $n=2 \Longrightarrow n+2$ is not prime, or only one out of three even numbers can be prime may vary or even trivial, but it is necessary.

But what do I know? Seems being right is for some users more important than teaching right.

 

[Math100]

I think you confuse right and stubborn.

The case of even $n$ requires one additional argument. Several are possible, but $n\equiv 0 \mod 2$ alone is not sufficient. The additional argument, $n>3$ or $n=2 \Longrightarrow n+2$ is not prime, or only one out of three even numbers can be prime may vary or even trivial, but it is necessary.

But what do I know? Seems being right is for some users more important than teaching right.

So is anything incorrect in my revised proof of post #13?

 

[pbuk]

So is anything incorrect in my revised proof of post #13?

Yes: the statement "Thus, there exists exactly one integer that is divisible by $3$" is incorrect (there are an infinite number of integers that are divisible by 3). If you mean

• Exactly one of $\{ n, n+1, n+2 \}$ must be divisible by $3$.

then that is what you should say.

It would also help if you didn't repeat that $n$ is an integer six times, or use phrases like "Let $n$ be an (even/odd/prime/composite) integer" instead of "Let $n$ be (even/odd/prime/composite)".

 

[fresh_42]

So is anything incorrect in my revised proof of post #13?

No, not at all. You simply should concentrate more on a strategy, a plan. The plan here is to find divisors $2$ or $3$. And the obstacle is, why do we need $n>3$? I would only add some comments.

Here is my view on it:

Suppose $n>3$ is an integer.
Now we consider two cases.
Case #1: Let $n$ be an even integer.
That is, $n=2k$ for $k>1$.
Then we have $n+2=2k+2$ and $n+4=2k+4$.
Note that all integers $n, n+2, n+4$ are even.
Thus, all integers $n, n+2, n+4$ are composites.

... since all primes $n>3$ are odd.

Case #2: Let $n$ be an odd integer.
That is, $n=2k+1$ for $k>1$.

For $k=1$ we get the triple $\{3,5,7\}$ of primes. Hence the statement requires $n>3$ which means $k>1$.

Then we have $n+2=2k+3$ and $n+4=2k+5$.
Suppose that $2k\equiv r \mod 3$.
Then we have $2k+1\equiv r+1 \mod 3$, $2k+3\equiv r \mod 3$ and
$2k+5\equiv r+2 \mod 3$.
Thus, there exists exactly one integer that is divisible by $3$.
Therefore, for $n>3$, the integers $n, n+2, n+4$ cannot all be prime.