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I'm learning math so this may seem obvious but its not to me.

In our other post titled

"Is this iteration involving primes known?"

Is this iteration involving primes known?

An iteration is defined that "stepped on primes" along the natural number line with steps of length some prime plus one.

The full details are given in the post above.

So considering the minimal step length the primes stepped upon (for a minimal step length) are thus

3, 11, 29, 61, 151, 331, 691, 1453, 2953, 5923, 11863, 23761, 47563, 95203, 190471, 381001, 762049, 1524277, 3048679, 6097417, 12194857, 24389767 and so on

Now consider the difference in consecutive terms in this sequence minus one. This is the "sequence of differences"

7,17,31,89,179,359,761 and so on up to 12194909 and the terms to infinity (but we must firstly prove that there are an infinite number of terms here) can be computed by iterating as described in the post referred to above.

So the above sequence is the minimal step lengths from prime to prime which by definition is prime.

Then the question is: can we prove this "sequence of differences" is infinite and prove why from 89 onward to infinity are the primes of the form 3a+2 (for some integer a) ?

It is implied from the above question that a proof is required that proves this sequences of differences is infinite. It was asked in the post

Is this iteration involving primes known?

if the sequence there was infinite and it obvious that it would follow then that the "sequences of differences" is infinite. So to proves these sequences are infinite requires proving an unproved statement about primes (this is given any prime $U$ find a prime $V$ such that $U+V+1$ is prime, $V$ can be minimal or not).

So we have not proved that this sequence is actually infinite (but suspect it is) see the post referred to above for more details.

So sadly this means the proof in the answer below is incorrect because it uses the property $x_n$ is infinite. It would be pleasantly surprising if the incorrect proof below was as short and simple as that, I'd love to know a correct answer.

The details why this proof is wrong is it assumes $x_n$ are infinite which we dont know to be true.The reason is because we do not have a proof of the following,if we are given any prime R, we must be able to prove that there exist another prime S, such that S is minimal, such that R+S+1 is prime. (Theres another unproved case when S isnt minimal) As far as I am aware there is no proof for this result. The question asks to prove if $x_n$ is infinite (not assume it is) and prove the form is 3a+2 of those integers in the infinite tail of the sequence

In addition to the above question the proof will also prove when the step length is not minimal. The non-minimal means the step length is now not the smallest possible prime. For example stepping on the prime 11 then a stepping by a non minimal length 41+1 to get to the prime 53. (The minimal step length is 17 this is why the term 29=11+17+1 appears above). This also means the proof below is incorrect as pointed out below

I don't understand why this proof got positive votes-maybe I'm missing something-so if you do vote the question up please leave the reason why.

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  • $\begingroup$ Note that after from 61, all numbers $n$ of the list are congruent to 1 modulo 3. Then, you have $n+1\equiv 2\mod 3$. Since every prime greater than 3 is congruent to 1 or 2 modulo 3, you can get the conclusion. $\endgroup$ – Ángel Valencia Feb 2 '16 at 21:57
  • $\begingroup$ @AngelValencia I see what you mean I think in the first list the they are all congurent to 1 modulo 3 thus in the second list the remainder is fixed when modulo 3. The question is then why all the primes in the first list are congruent to 1 modulo 3? Is this easy to see? $\endgroup$ – Karlie Kloss Feb 2 '16 at 22:03
  • $\begingroup$ It because I am personally interested in this problem. I wasnt aware that the problem I posed was impossible to understand if that is what you meant. $\endgroup$ – Karlie Kloss Feb 2 '16 at 22:06
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    $\begingroup$ Please avoid having a meaningless discussions in the comments. If you don't understand the problem and you think it's a waste of time to look into it, then move on. Let them have some fun with this sequence, please. $\endgroup$ – vrugtehagel Feb 2 '16 at 22:13
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    $\begingroup$ @user1952009, you should know that showing that 8 and 9 are the unique consecutive integer powers was enough to publish a paper (cf. J. Reine Angew. Math. 572 (2004), 167-195). There aren't un-interesting questions; every question lead us to discover new properties of prime numbers and math in general. $\endgroup$ – Ángel Valencia Feb 3 '16 at 3:07
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I'd like to define your sequence in a, in my opinion, cleaner way (although very similar). We define $x_1=3$. We now define $x_n,y_n$ by the following properties.

$$x_n,y_n \text{ prime}$$ $$\text{the smallest prime solution of } x_{n+1}=x_n+a+1 \text{ is }a=y_n$$ We'll assume the sequences are infinitely long.


Now we'll do some induction to prove $$x_n\equiv 1\mod 3$$ for $n\geq 4$. The base case is $x_4=61\equiv 1\mod 3$. Now assume $x_n\equiv 1\mod 3$ is a prime. Now we have a prime $y_n$ (actually the smallest, but that's not relevant to the proof) such that $x_{n+1}=x_n+y_n+1$ is prime. Since primes are not divisible by $3$ (that is, not $0\mod 3$), we know $$x_n+y_n+1\equiv y_n+2\not\equiv 0\mod 3$$ so $y_n\not\equiv 1\mod 3$. Since $y_n$ is prime, it cannot be divisible by $3$ - so $y_n\equiv 2\mod 3$ (the only option left). This also means that $x_n+y_n+1\equiv 1+2+1\equiv 1\mod 3$, so that $x_{n+1}\equiv 1\mod 3$.
Now we proved that $x_n\equiv 1\mod 3$ for all $n\geq 4$. the argument above, that showed $y_n\equiv 2\mod 3$, now holds for all $n\geq 4$.

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  • $\begingroup$ It's induction. Once it's at $1$, it'll stay $1$. The base case is $x=61$, so before that, it doesn't work. $\endgroup$ – vrugtehagel Feb 2 '16 at 22:44
  • $\begingroup$ I'm proving here that $x\equiv 1\mod 3$ for all $x\geq 61$. That's equivalent with $y\equiv 2\mod 3$. Could you specify the part that you don't understand, so that I can improve my answer? $\endgroup$ – vrugtehagel Feb 2 '16 at 22:56
  • $\begingroup$ I edited my answer. Please let me know if I'm still being unclear, and if I am, what part I'm being unclear at. $\endgroup$ – vrugtehagel Feb 2 '16 at 23:19
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – vrugtehagel Feb 2 '16 at 23:30
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    $\begingroup$ Becuase this proof assumes $x_n$ are infinite which we dont know to be true.The reason is because we do not have a proof of the following,if we are given any prime R, we must be able to prove that there exist another prime S, such that S is minimal, such that R+S+1 is prime. (Theres another unproved case when S isnt minimal) As far as I am aware there is no proof for this result. The question asks to prove if $x_n$ is infinite (not assume it is) and prove the form is 3a+2 of those int the infinite tail of the sequence $\endgroup$ – Karlie Kloss Feb 3 '16 at 4:16
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A more general problem is the following: if $x$ and $y$ are two prime numbers such that $$x<y,\qquad x\equiv1\mod3\qquad\text{and}\qquad x+y+1\text{ is prime},$$ then $y\equiv2\mod3$. In fact, by hypothesis we have $$x+y+1\equiv y+2\mod3.$$ Now, since $x+y+1$ is prime, we have $x+y+1\equiv1$ or $2\mod 3$. This implies $$y\equiv-1\text{ or }0\mod 3.$$ The condition $y\equiv0\mod3$ implies $y=3$, which is discarded because there aren't prime numbers less than $3$ which are $\equiv1\mod3$; thus $y\equiv-1\equiv2\mod3$.

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