Condition in Type of Prime Factors of Consecutive Integers

Question

We define a odd-prime $p$ as $i$-type prime if $p \equiv - i \pmod q$ where $ 1 \leq i \leq q-1$ (see similar definition on page 24, CHAPTER 2, of the book "Summing It Up" by Avner Ash andRobert Gross, 2016), here the given integer $q$ is fixed, like residue class.

If $s $ has primes of all $i$-type prime,

can we say (deterministically predict) for which $j$-type prime (where $1 \leq j \leq q-1$) can't divide $s+1$?

For example, $ q=4, i=\{1, 3\}, s=5 \times 7 =35,$ now, $5= p_1, 7=p_3$ both divides $s$, but $s+1 =35+1=2^2 \times 3^2$, and $3=p_3$, is there any proposition (theorem/lemma), result in book or journal, algorithm or method to predict, that $p_1$ is not going to be a factor of $s+1$ (in this case, this is just an example).

What is the related topics to this problem? Please comment anything related to the problem.

Please consider the NON-TRIVIAL cases.

EDIT:

Page 24, CHAPTER 2, of the book "Summing It Up" by Avner Ash andRobert Gross, 2016 -

Answer 1

Your question is quite broad and may not have a general result for each randomly picked integer $s$; however, if we desire certain classes of such integers, then the answer is yes. My answer can be derived from the following classical/folklore result attributed to Euler:

Let $\Phi_q$ be the $q$th cyclotomic polynomial, then the only prime factors $p$ of $\Phi_q(n)$, for any integer $n$, are $p\equiv 1\mod q$ or $p|q$.

This result gives the classical Euclid-style proof that there are infinitely many prime numbers $\equiv 1\mod q$. It follows that if $s=\Phi_q(n)-1$, the the prime factors of $s+1$ are now restricted. In the especial case when $q=2^{m+1}$, for some natural number $m$, one has $$\Phi_{2^{m+1}}(n)=n^{2^m}+1\,$$ which means that you can choose $n$ to be composed of primes in the same arbitrary residue class modulo $2^{m+1}$ and set $s= n^{2^m}$ and you immediately obtain that $s+1$ won’t have prime factors, besides possibly $2$, that are not in the trivial residue class modulo $2^{m+1}$.

Answer 2

The problem

I will start by reformulating a bit, so people will understand better what the problem is about:

Given an integer $n$, and knowing the factorization (in prime factors) of $m$, can we predict what residues will the prime factors of $m+1$ give$\pmod{n}$?

A rather easy case

I will firstly discuss the case when $n=4$, just as an introductory approach for the general case. The answer to your question is that we can sometimes predict what residues the prime factors of $m+1$ will give $\pmod{4}$. In what ways can we do so?

• For example, if $m\equiv 2\pmod{4}$, we can for sure tell that $m+1$ is divisible by a prime $\equiv 3\pmod{4}$
• However, if $n\equiv0\pmod{4}$, you cannot tell if $n$ is divisible or not by a simple modular analysisif a prime $\equiv 1$ or $\equiv3\pmod{4}$. The same goes for $n\equiv\pmod{4}$ and $n\equiv\pmod{4}$.

So this wasn't very satisfactory. Lets use some stronger methods, some theoremes (for this approach, analyzing$\pmod{4}$ is especially attractive). Here is what we can deduce:

• For example, if $m=k^2$, then using some quadratic reciprocity we can for sure say there is no prime $\equiv 3\pmod{4}$ which divides $m^2+1$. More generally, if $m=a^2+b^2-1$ such that no prime $\equiv 3\pmod{4}$ divides $ab$, then again, we can say there is no prime $\equiv 3\pmod{4}$ which divides $a^2+b^2$.
• As another example, if $m=a^{\phi(b)}-2$ and $gcd(a;b)=1$ , we can conveniently use Euler's theorem to see that $b$ divides $m+1$ and from there deduce whether some primes $\equiv 3$ or $\equiv 1\pmod{4}$ divide $m+1$.

This wasn't too satisfactory either. We can indeed find very many forms of $m$ for which we can deduce the residues of some prime factors of $m+1$ using some theoremes, but those cases are (as I said before) unsatisfactory. They are few, too specific and... for the generalized case, most of the approcahes stop to work ( for example the quadratic reciprocity method and the modular arithmetic method).

So I think there is only one approach left, the most powerful one, which isn't restricted by either $n$ or the form of $m$, which I will use for $n=$ and then try to generalize for any $n$: probability.

Yes, I think we can "deduce" the residues of the prime factors of $m+1$ using some probabilistic arguments. Of course, we cannot for sure find what residues the prime factors of $m+1$ will give, but I don't think there are any arguments stronger than the probabilistic interpretation.

Edit: Sorry, Andrew, in the past days I have felp worse and I am not capable of concentrating. I rested. I am sorry I couldn't help you. Farewell.