Prime Triple
A , A + 2 , A + 4
If A > 3 , can all three numbers above be prime?
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9 solutions
Discussions for this problem are now closed
Can you also give a proof of this statement?
If a is odd then a mod 3 gives 1 or 2. In the first case a+2 mod 3=0. Otherwise a+4 mod 3 =0
9 is odd but 9 m o d 3 = 0 ... if a is also prime then your statement is true.
If a is odd then a m o d 3 gives 0 , 1 or 2 . If it's 0 then a is a multiple of 3. If it's 1, then a + 2 m o d 3 ≡ 3 ≡ 0 . If it's 2 then a + 4 m o d 3 ≡ 6 ≡ 0 , so, respectively a + 2 and a + 4 would be multiples of 3.
@David Molano – if a is prime and larger than 3 then a is alway odd number, and the remainder is 1 or 2
@Hoai-Thu Vuong – If you would read the rest of the comments...
@David Molano – It's already given that a is prime - otherwise the starting condition is false and we can stop there. Riccardo's proof is simplest and best.
@Yury Itskovich – Yes. However the same style of proof works to prove something more general, that the same holds even if a is not a prime. And even if a isn't odd.
@David Molano – Riccardo's statement begins with "If A is prime". If A isn't prime, then there is nothing to prove as the starting condition is false.
@Yury Itskovich – That's also true, but not what I'm discussing. He uses this "Lemma":
"If A is prime then A+2 or A+4 is a multiple of 3".
Then, to prove it he uses this:
"If a is odd then a mod 3 gives 1 or 2. In the first case a+2 mod 3=0. Otherwise a+4 mod 3 =0"
And here I'm supposing the lower case "a" is not the same as the upper case "A". If they are the same it's done. If they're not the same then we must also think of the case a mod 3=0. And in the proof the hypothesis "a is odd" is not used, because it's not really linked to "a mod 3 gives 1 or 2".
And in any case, I say this can be said as the corollary of a more general "Lemma":
"If a is any integer, then one of a , a + 2 , a + 4 is a multiple of 3".
@David Molano – If A mod 3 = 0 and A > 3, then A is not prime because it's divisible by 3.
@Alex Bean – Did you read everything?
@David Molano – Yes. When he says a, he means A. Dunno why he uses "a is odd". I tried to clarify because you seemed confused.
@Alex Bean – I have it clear. What I am saying is that even if he refers to A as a, the same proof works to prove something much more general.
@David Molano – OK, yeah, I misunderstood you.
Numbers A, B, C have pairwise different remainders modulo 3, because their pairwise differences are not multiple of 3.
There are exactly 3 remainders modulo 3, so some of A, B, C have remainder 0.
One of these three numbers is a multiple of 3: A, A+1, and A+2 (because every three consecutive numbers, there is a multiple of 3, as in 3, 6, 9, 12, etc.). If A+1 is a multiple of 3, A+4 is also a multiple of 3.
Not really, 2 is prime, therefore; Would 4 and 6 both be multiples of 3 5; 7 & 9 7: 9 & 11 Etc.. XD, just kidding add "either" after "then", it was a bit misleading :P
If A is a prime greater than 3. (Which is given)
If A=5, then B=7 and C=11, and all are prime. Oh, I see my mistake. The last one is A+4, not B+4....
What if A=11 B=13 C=17 all are prime
C has to be A +4 =15 not prime
You made the same mistake I made. The last number is A+4, not B+4.....
A+2 cannot be even, or else it would be divisible by 2.
If A+2 is odd, A, A+2, and A+4 are consecutive odd numbers. Out of 3 consecutive odd numbers, one of them is always divisible by 3. (Except 1, 3, 5 of course. But A>3.) So all three cannot be prime.
I think you meant except 3, 5, 7? 1 is not prime.
A is greater than 3 (given), not greater or equal to.
- Decompose into 3 cases:
- A=3k: not possible since A is not prime
- A=3k+1: then A+2=3k+3 is not prime
- A=3k+2: then A+4=3k+6 is not prime
- No matter what A is, A,A+2 and A+4 cannot be all primes.
Consider A as a number mod 3. Then there are only 3 cases to consider: A= 0 mod 3
A = 1 mod 3
A = 2 mod 3.
In the first case, A is itself not prime. So it does not satisfy the relation.
In the second case A = 1 mod 3 implies A+2 = 0 mod 3 (0 = 3 mod 3 ) which is not prime. So it also does not satisfy the relation.
Lastly, A= 2 mod 3 implies A+4 = 0 mod 3 ( 6 = 0 mod 3) which is not prime.
In each case, one of the numbers ends up being a multiple of three. Therefore, it is impossible.
I did this a very different way, my mathematical knowledge is a bit basic when it comes to number theory so I have to use patterns. First I noticed that any solution would be of the form:
3+D,5+D,7+D.
(The starting point of 3,5,7 is not really relevant only that they are each two apart.)
Next I looked for the value of D for each new solution for each A,A+2,A+4.
Each column is A,A+2,A+4, and each row is the value of D for the next prime.
2 ,2 ,4
4 ,6 ,6
8 ,8 ,10
10,12,12
There is a very clear pattern here and it can be seen that for any two to be prime the third will always be non-prime. However, this is assuming that the pattern will go on forever which i can not prove.
What is the reason for this cool pattern? I tried extending it backwards but it didn't make much sense. If a different starting point to 2,5,7 say 4,6,8 is chosen, I find exactly the same pattern.
The question is asking for 3 consecutive primes which have a difference of 2, of which 3,5,7 is the only example. Since A is to be bigger than 3, A, A+2, and A+4 cannot all be prime.
Any prime numbers greater than 3 will satisfy either (6n-1) or (6n+1).we cannot have 3 consecutive odd numbers that satisfy (6n-1) or (6n+1) and greater than 3. so "not possible".
If A is a multiple of 3, plus 1, then A+2 is a multiple of 3. If A is a multiple of 3, plus 2, then A+4 is a multiple of 3. There are no alternative scenarios.
Coprimes only exist in pairs, not triplets. Ex: 5 and 7, 11 and 13, etc
Since 3 is a prime, anything divisible by three cannot be a prime. So if x is a prime and x + 2 is a prime, then x + 4 cannot be. (2 + 4 is 6, a multiple of 3).
If A > 3 is prime, then it leaves a remainder of either 1 or 2 when divided by 3. Thus, either A + 2 or A + 4 must be a multiple of 3.