[SOLVED] 3, 5, 7 primes

**dwsmith** · June 3rd 2010, 09:12 AM

Prove that 3, 5, 7 are the only three consecutive odd integers that are primes.

How to start?

**undefined** · June 3rd 2010, 09:20 AM

Originally Posted by dwsmith

Prove that 3, 5, 7 are the only three consecutive odd integers that are primes.

How to start?

Consider congruence, mod 3. For any three consecutive odd integers, exactly one of them will be congruent to 0 (mod 3). If it is greater than 3, then it cannot be prime.

**dwsmith** · June 3rd 2010, 09:22 AM

Originally Posted by undefined

Consider congruence, mod 3. For any three consecutive odd integers, exactly one of them will be congruent to 0 (mod 3). If it is greater than 3, then it cannot be prime.

I understand modulo math but I am not sure how it would be applied. Can you elaborate some more?

**undefined** · June 3rd 2010, 09:27 AM

Originally Posted by dwsmith

I understand modulo math but I am not sure how it would be applied. Can you elaborate some more?

Sure.

Three consecutive odd integers can be written n, n + 2, n + 4. We know n is of the form n = 2k + 1, but this is not important for our proof.

n is either congruent to 0, 1, or 2 (mod 3). This makes 3 cases.

Suppose n is congruent to 0 (mod 3). Then it is a multiple of 3. If it is not 3, then it is not prime, and we are done.

Suppose n is congruent to 1 (mod 3). Then $n + 2 \equiv 1 + 2 \equiv 3 \equiv 0\ (\text{mod}\ 3)$ . And we are done.

Suppose n is congruent to 2 (mod 3). Then $n + 4 \equiv 2 + 4 \equiv 6 \equiv 0\ (\text{mod}\ 3)$ . And we are done.

**dwsmith** · June 3rd 2010, 09:41 AM

Originally Posted by undefined

Sure.

Three consecutive odd integers can be written n, n + 2, n + 4. We know n is of the form n = 2k + 1, but this is not important for our proof.

n is either congruent to 0, 1, or 2 (mod 3). This makes 3 cases.

Suppose n is congruent to 0 (mod 3). Then it is a multiple of 3. If it is not 3, then it is not prime, and we are done.

Looking at this first case:

$n\in\mathbb{Z}^+, \ n=2k+1$ and three consecutive odd integers would be of the form $n, n+2, n+4$

Suppose $n\equiv 0 \ (\text{mod} \ 3)$

$n=$ { $3,6,9,12,...$ }

This is understandable but then don't I have to then show 3 would be the only odd prime out of the set that satisfies this congruence?

**undefined** · June 3rd 2010, 09:45 AM

Originally Posted by dwsmith

Looking at this first case:

$n\in\mathbb{Z}^+, \ n=2k+1$ and three consecutive odd integers would be of the form $n, n+2, n+4$

Suppose $n\equiv 0 \ (\text{mod} \ 3)$

$n=$ { $3,6,9,12,...$ }

This is understandable but then don't I have to then show 3 would be the only odd prime out of the set that satisfies this congruence?

There's not much to show. A multiple of 3 is of the form n = 3m. If m > 1, then n is composite (it has factors m and 3, among possibly others).

**dwsmith** · June 3rd 2010, 09:52 AM

Originally Posted by undefined

Suppose n is congruent to 1 (mod 3). Then $n + 2 \equiv 1 + 2 \equiv 3 \equiv 0\ (\text{mod}\ 3)$ . And we are done.

If we are saying $n\equiv 1\ (\text{mod}\ 3)$ , why did you immediately sub in $n+2$ .

**undefined** · June 3rd 2010, 09:57 AM

Originally Posted by dwsmith

If we are saying $n\equiv 1\ (\text{mod}\ 3)$ , why did you immediately sub in $n+2$ .

I was looking for a member of {n, n+2, n+4} that was divisible by 3.

**ANDS!** · June 3rd 2010, 10:05 AM

This was a problem we did, the reasoning is the same but does (explicitly) use modulus and congruence to prove:

As stated, three consecutive integers are of the form: n, n+2, n+4. The first integer is divisible by three with a remainder of 0, 1, or 2. 0 is obviously out.

If n has a remainder of 1:

$n = 3k+1 \Rightarrow n+2 = (3k+1)+2 \Rightarrow n+2 = 3(k+1);$ which is a multiple of 3 and thus not a prime number.

If n has a remainder of 2:
$n = 3k+2 \Rightarrow n+4 = (3k+2)+4 \Rightarrow n+4 = 3(k+2);$ which is a multiple of 3 and thus not a prime number.

Obviously not as succinct and to the point as Undefined's proof, but if you are in the same spot I was in when this question came up, this was what our Professor expected us to know.

**dwsmith** · June 3rd 2010, 10:16 AM

Originally Posted by undefined

I was looking for a member of {n, n+2, n+4} that was divisible by 3.

Let $n\in\mathbb{Z}^+ \ \text{such that} \ n=2m+1$ ; thus, 3 consecutive odd $\mathbb{Z}$ would be of the form $S=$ { $n, n+2, n+4$ }.

Case 1: $S\equiv\ 0\ (\text{mod}\ 3)$
Let $n\equiv\ 0\ (\text{mod}\ 3)$ .
$n=$ { $3,6,9,...$ }; therefore, $n=3$ or $n$ isn't prime, since if $n=3m$ and $m>1$ , then $n$ is composite.

Case 2: $S\equiv\ 1\ (\text{mod}\ 3)$
Let $n+2\equiv\ 1\ (\text{mod}\ 3)$ .
$n=$ { $2,5,8,...$ }.

Case 3: $S\equiv\ 2\ (\text{mod}\ 3)$
Let $n+4\equiv\ 2\ (\text{mod}\ 3)$ .
$n=$ { $1,4,7,...$ }.

I am not sure now how to tie in that it follows $n=5$ in case 2 and $n=7$ in case 3.

Obviously since we are looking for 3 consecutive odd primes it must be those.

Also, how would I then show that these are the only 3 consecutive?

**ANDS!** · June 3rd 2010, 10:26 AM

Also, how would I then show that these are the only 3 consecutive?

By exhaustion of those two cases. If N is congruent to 1(mod 3), then N+2 is congruent to 0(mod3); and if N is congruent to 2(mod3), then N+4 is congruent to 0(mod3).

Also 1(mod3) should be {1, 4, 7. . .} and 2(mod3) should be {2, 5, 8. . .}.

**dwsmith** · June 3rd 2010, 10:30 AM

Originally Posted by ANDS!

By exhaustion of those two cases. If N is congruent to 1(mod 3), then N+2 is congruent to 0(mod3); and if N is congruent to 2(mod3), then N+4 is congruent to 0(mod3).

Also 1(mod3) should be {1, 4, 7. . .} and 2(mod3) should be {2, 5, 8. . .}.

If $n+2\equiv 1 \ (\text{mod} \ 3)$ , then $n\neq 1$ since $1+2=3\not\equiv 1 \ (\text{mod} \ 3)$

**undefined** · June 3rd 2010, 10:31 AM

Originally Posted by dwsmith

Let $n\in\mathbb{Z}^+ \ \text{such that} \ n=2m+1$ ; thus, 3 consecutive odd $\mathbb{Z}$ would be of the form $S=$ { $n, n+2, n+4$ }.

Case 1: $S\equiv\ 0\ (\text{mod}\ 3)$
Let $n\equiv\ 0\ (\text{mod}\ 3)$ .
$n=$ { $3,6,9,...$ }; therefore, $n=3$ or $n$ isn't prime, since if $n=3m$ and $m>1$ , then $n$ is composite.

Case 2: $S\equiv\ 1\ (\text{mod}\ 3)$
Let $n+2\equiv\ 1\ (\text{mod}\ 3)$ .
$n=$ { $2,5,8,...$ }.

Case 3: $S\equiv\ 2\ (\text{mod}\ 3)$
Let $n+4\equiv\ 2\ (\text{mod}\ 3)$ .
$n=$ { $1,4,7,...$ }.

I am not sure now how to tie in that it follows $n=5$ in case 2 and $n=7$ in case 3.

Obviously since we are looking for 3 consecutive odd primes it must be those.

Also, how would I then show that these are the only 3 consecutive?

Your proof has some problems, such as confusion between sets and integers. I would avoid setting an integer equal to a set unless you're doing analysis and defining the integers, etc. I would also avoid the shorthand "3 consecutive odd $\mathbb{Z}$ ." What you really mean is "3 consecutive odd members of $\mathbb{Z}$ " but it is easier to just write "3 consecutive odd integers."

It might help you to let n > 3. (The case n = 1 is easily dealt with separately.)

**undefined** · June 3rd 2010, 10:34 AM

Originally Posted by dwsmith

If $n+2\equiv 1 \ (\text{mod} \ 3)$ , then $n\neq 1$ since $1+2=3\not\equiv 1 \ (\text{mod} \ 3)$

It is easiest to make the three cases:

Case 1: $n \equiv 0\ (\text{mod}\ 3)$

Case 2: $n \equiv 1\ (\text{mod}\ 3)$

Case 3: $n \equiv 2\ (\text{mod}\ 3)$

**dwsmith** · June 3rd 2010, 10:35 AM

Originally Posted by undefined

It is easiest to make the three cases:

Case 1: $n \equiv 0\ (\text{mod}\ 3)$

Case 2: $n \equiv 1\ (\text{mod}\ 3)$

Case 3: $n \equiv 2\ (\text{mod}\ 3)$

I don't like the confusion of n and then having n+2 and n+4. That is why I said S is equiv and then choose the appropriate element of S.

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