Prove that no "prime triplet" exists after 3,5,7

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The discussion centers on proving that the only prime triplet is 3, 5, 7, with no other triplets existing. The approach involves analyzing the properties of numbers in relation to multiples of 3. It is established that any prime triplet must include a number that is a multiple of 3, which disqualifies subsequent triplets from being entirely prime. The proof demonstrates that for any integer p in the forms of 3j, 3j + 1, or 3j + 2, at least one of p, p + 2, or p + 4 will not be prime. Thus, it is concluded that no prime triplets exist beyond 3, 5, 7.
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So,

Prime triplet: if p is prime, then p+2 and p+4 are also prime.
Prove that 3,5,7 is the only prime triplet.

I have figured out that

p = 2k+1, k > 1

then p+2 = 2k + 3 and p + 4 = 2k+5.

I figure there will be three cases to prove i.e.

p (prime), p + 2 (prime), p + 4 (not prime)

p (prime), p + 2 (not prime), p + 4 (prime)

p (prime), p + 2 (not prime), p +4 (not prime)

Not sure how to continue. The solution is in the book, but I just want a little hint in the right direction...and I hoped writing it out again would give me some hint. I am just starting with number theory and haven't done many proofs in my life so this feels a little tough.
 
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Hint #1: You're thinking too hard. This is an easy problem.
Hint #2: How many members are in a triplet?
 
Hint #3: select a few triplets, do you see anything common about them?
 
And since the OP hasn't replied yet,

Hint #4: Every such triplet (including 3,5,7) includes one member that is a multiple of what number?
 
Your first hint is to look at a group of them and then understand what is special about the anomaly.

3, 5, 7 works
5, 7, 9 doesn't
7, 9, 11 doesn't
9, 11, 13 doesn't
11, 13, 15 doesn't

What do all of these have? How does 3, ,5, 7 get away with it? Can you prove that any triplet will have the same problem?
 
Hopefully that's enough hints to get OP started.

Note how I refrained myself from posting another hint.
 
Borek said:
Hopefully that's enough hints to get OP started.

Note how I refrained myself from posting another hint.
I took that as a hint to me to also refrain thusly !
 
haha thanks so much everyone!
Every triplet has a multiple of 3. The first triplet survives this because the multiple is 3 itself, and 3 is prime. However, every triplet after the first will contain a multiple of 3 which by the definition of prime numbers means one element is not prime.

So let's assume p = 3j where j is a positive integer. Then every integer can be formed by 3j + 1 or 3j + 2.

Then for the first case j = 1 we have p = 3, p + 2 = 5 and p + 4 = 7. No contradiction since we have a prime triplet.

For the next case 3j + 1 we get p = 3j + 1, p + 2 = 3j + 3 = 3 (j + 1), contradiction since it is a multiple of 3.
So we can't have a triplets where p is this form.

Finally, for p = 3j + 2 we get p = 3j + 2, p + 2 = 3j + 4, and p + 4 = 3j + 6 = 3 (j + 2) --> contradiction multiple of 3. Thus, we cannot have prime triplet of this form.

Therefore no prime triplets exist after 3, 5, 7

Thanks all!
 
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