Question

Dear CHAMPIONS !
- samthing new about primes
5 - 7
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5+7=12 +/- 1 = 11 - 13
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11+13=24 x 3 = 72 +/- 1 = 71 - 73
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72+73=144 x 3 = 432 +/- 1 = 431 --- 433
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431+433=864 x 3 = 2592 +/- 1 =
= 2591---2593
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- so than you see these all twins what have resulted begin this line with 5 - 7
every resulted twins has members ending in 1 and 3 - so not is this very very interesting ?

Answer 1

- so than we like continue this line - what we get ? ---

- than we begin it by 17 --- 19
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17+19=36 x3 = 108+/- 1 = 107 --- 109
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107 +109 = 216 x3 = 648+/-1 = 647 --- 649
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- so than you see them started by 17 and 19 every resulted twins has mambers ended in 7 and 9 - not is interesting too ?

29 --- 31
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29+31 = 60 +/-1 = 59 --- 61
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59+61=120 x3 = 360+/-1= 359 --- 361
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- so every two twins has members ending in 59 and 61 - how much interesting it these ?

Answer 2

@ganeshie8
@ParthKohli
@satellite73
@zepdrix
@mathmale
@saifoo.khan

Answer 3

@phi
@sleepyjess
@SolomonZelman
@Empty
@Kainui

Answer 4

Well, you have a proof in front of you. Which is it.

1) Will it ALWAYS find Twin Primes?
2) Will it find ALL twin Primes?

Go!

Answer 5

ok. i will try it @tkhunny how you have said me above

- we know that every twins has a form of 6n+/-1 - yes ?
started by 5 - 7 what are 6*1+/-1= 5 -- 7
- and continue it how we say the complet math induction proof method the above line let be the first step what always is the case for n=1 - so this is true
- the second step say that we suppose this is true for n=k what mean 6k+/-1

and in the 3rd step we need prove it for k=k+1 - what will be in the form of 6(k+1)+/-1

- let we see it what we get ???

6(k+1)+/- 1 = (6k+6) +/- 1 = 6k +5 and 6k+7 so for k=1 we get 6*1+5=11 and 6*1+7 = 13

so we have got 11 -- 13 what are twins

- so we can saying ,,q.e.d"

- opinions please !!! - thank you very much

Answer 6

I'm a bit confused about your procedure and the facts you find interesting, but this may shed light on the behavior that I believe is of interest.

First, from what I see, you take \(p_n \) and \(p_m\) and then find \(p_n+p_m\) where \(p_n, p_m > 2\). Then, sometimes, you form your new numbers by \( 3(p_n + p_m) \pm 1\), which must take the form \(6k \pm 1\) and therefore takes the form of all twin primes.

Second fact was the ending digits. If we look at \(p_n\) and \(p_m\) where \( p_n \equiv a \bmod(10) \) and \( p_m \equiv b \bmod (10)\) then:
\[3(p_n + p_m) = 3p_n + 3p_m \equiv 3a + 3b \bmod(10) \]
And since \(b-a = 2 \implies b = a+ 2\) then:
\[3(a+b) -1 = 3(2a+2) -1 = 6a + 5 \] and \[ 3(a+b) + 1 = 6b+5\]

We wish to show that \(6a+5 \equiv a \bmod(10)\) when \(a\) is odd.

Since we are only concerned with the least significant digit, we are in \(\mathbb{Z}_{10}\) . Every odd except \(a=5\) is a generator since they are relatively prime to \(10\). That is \(11a = a\) in this group if \(a \in \{1,3,7,9\}\).
The fact that \(5\) generates a sub group allows us to finish the proof. That is \(2(5) = 0 \) (or identity element \(e\)). But since the \(a\) shown above also give \(10a = 0 = e\) then \(10(a) = 2(5)\) which implies \(5 = 5(a)\).

From this, we get that \(6a + 5 = 6a + 5a = 11a =a\) where \(a\) is odd. That is, given your procedure, we should expect that they always produce two numbers with the same one's digits.

I believe similar reasoning can show it for \(\mathbb{Z}_{100}\) since we are guaranteed \(a, b\) will be relatively prime and therefore generators.

Answer 7

In haste, there is a false implication. While it is true that 5a=5 for the elements a in \( \mathbb{Z_{10}} \), it is not implied from the sole fact alone, but requires that the subgroup generated by 5 is of order 2.

In \(\mathbb{ Z}_{100}\), 5 generates a group of order 20 which implies \(100p_n=20(5) \) that there will be a \(kp_n=5\) such that \(5|k\). Since we are looking for twin primes with this property, we want to find a prime \( p_n \) where \(6p_n+5=p_n\) which means we need to find a \(kp_n=5\) where \(k=100−5=95\). There are only two relatively prime numbers that satisfy this property in \(\mathbb{ Z}_{100}\) and they are 19, 59 and only two relatively prime numbers have the property \(k p_m + 95 = p_m\)

59, 61 are the only pair of twin primes between 1-100 that have the property seen above. If you follow your process with 19, 21 since if \(6p_n + 5 \equiv p_n \bmod(100)\) then \(6(p_n+2) + 5 \equiv p_n + 2 \bmod(100)\). Those relatively prime twin primes will have the same property (but those are the only 2 in that range).