How many primes should I look at?

Relevant wiki: Division Algorithm

All numbers can be represented as 3k, 3k+1 or 3k+2 where k is a positive integer.
If X can be represented as 3k, X will not be prime as 3k is a multiple of 3. 3 is an exception as 3 itself is a prime number.
If X can be represented as 3k+1, X+2 is not prime as (3k+1)+2= 3k+3= 3(k+1) is a multiple of 3.
If X can be represented as 3k+2, X+4 is not prime as (3k+2)+4=3k+6=3(k+2) is a multiple of 3.

Relevant wiki: Prime Numbers

One out of $3$ consecutive odd numbers will be divisible by 3. The only prime that is divisible by $3$ is 3. So if is odd and greater than $3$ , this is not possible.

Now we only have to check $X=2$ . For $X=2$ , then $4$ and $6$ are not prime, so this does not work.

Therefore, $3$ is the only $X$ where all three are prime.

suppose, X=odd, because without 2, no even number is prime.........[so, even numbers are irrelevant for this case.]

now, if, $X=odd$ , then,in every 3 or 4 numbers, there will appear at least a digit which is a factor of $3,5,7$ etc

so, this is clear that,this condition is not true for all case(any other X)

Relevant wiki: Pythagorean Theorem

By applying the Pythagorean Theorem.

${ (x) }^{ 2 }\quad +\quad { (x+2) }^{ 2 }\quad =\quad { (x+6) }^{ 2 }$

You get $x$ = 6. Therefore, the answer is no .

No. For any three consecutive odd numbers, exactly one will be a multiple of three. For all other X; X is of the form 3a, 3a-1, or 3a+1. Trivial therefore to see that one of X, X+2 and X+4 is therefore divisible by 3. Hence "No." One of XX, X+2X+2, and X+4X+4 must be divisible by 3. Because of this, all three can be prime only if one of them is 3.

X=3X=3 gives 3, 5, 7 which works.

X=1X=1 gives 1, 3, 5 does not work because 1 is not considered a prime.

X=−1X=−1 gives the even more ridiculous -1, 1, 3.

So, 3, 5, 7 is the only such triplet, although there are prime triplets of the form XX, X+2X+2, X+6X+6 or XX, X+4X+4, X+6X+6. No, this is not true for any other x.

3+6n, where n is any natural number, cannot be prime since it obviously will be divisible by 3.

Let’s say x-2 is an odd number divisible by 3. x and x+2 will not be divisible by 3 so depending on what x is, they could both be prime. However x+4 will be 6 more than a multiple of 3, meaning that it must also be a multiple of 3 and therefore cannot be prime.

If you remove the x+4 requirement, then there are an infinite amount of possible x values satisfying the condition that x and x+2 are both prime. Google twin prime conjecture if you want more info on that. No, if X can be any other number, just X could be a non-prime number. The question says “true for any other X”, so any exception disproves the question. For example, if X = 6, it is divisible by 1, 2, 3 and 6, making it not a prime number. X+2=8, which is divisible by 1, 2, 4 and 8, and also not a prime number. Etc… As a little thinking will show, all primes beyond 3 are of the form 6n±1, where n is a positive integer. For example, if n is 1 then 5 and 7 are candidate primes, fulfilling the 3, 3+2 and 3+4 rule.

If n = 2, 11 and 13 are candidates. But the x then is 9, not prime.

n = 3 gives 17 and 19 but x is 15, not prime.

n = 4 gives 23 and 25m x is 21, not prime.

Notice the progression? x is 6n-3, easily proved. But 6n-3 isn’t a prime, it’s a multiple of 3.

6n-3 = 3(2n-1).

So NO the case of 3, 5, 7 is the only one that follows the rule required.

(Woo, I’ve done a number-theory proof! Just a few lemmas to fill in…)

"X, X+2,X+4 where X is prime other than 3"

gn is the difference between to successive Prime Numbers. The n-th and n+1-th prime.

As in gn= pn+1- pn

Example: pn = X and pn+1 = X+2 gn= X+2-X =2 - > gn = 5 - 3 = 2

It follows from the condition above that

gn = pn+1 - pn = 2 in two consecutive cases besides for X=3.

gn = 2 consecutively exist only between primes 3,5,7 which can be described as X, X+2, X+4.

Therefore there is no other prime for which the condition X, X+2,X+4 holds up.

One of them must be divisible by 3! (To be slightly more rigorous x + 4 = x + 1 (modulo 3), and one of x, x + 1 and x + 2 must be divisible by 3. Regards, David

First of all to recognise the prime number we have to use rules, first one the number must be greater than 1 and it has only two factors one and itself.

We know that 1 is not a prime number, so taking X=1 won't work. We also can't take X as an even number as

$EVEN$ $+$ $EVEN=EVEN$ and any even no. greater than 2 isn't prime.

Now for X>3 and odd, one out of every 3 consecutive odd numbers is divisible by 3.

SO we're screwed anyways