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Let $a$ be an integer. Suppose there exists infinitely many primes $\{p_i\}_{i \in \mathbb N}$ such that $a \equiv 1 \pmod{p_i}$, then it's clear that $a$ has to equal to $1$.

Now, suppose there exists infinity prime $p_i$, such that there exist $n_i$ with $a \equiv 2^{n_i} \pmod{p_i}$. Does this imply that $a$ needs to equal to $2^n$ for some integer $n$?

I am a bit struggling to show this because the power $\{n_i\}_{i \in \mathbb N}$ is not bounded.

Any hint or idea would be really appreciated.

Arturo Magidin
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ghc1997
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    Without currently having a rigorous proof , I am pretty sure that there are inifnite many primes dividing $2^n-3$ for some suitable $n$. – Peter May 14 '24 at 18:47
  • Hint: as here, mimic Euclid's proof of infinitely many primes to show that there are infinitely many primes $,p,$ where $,a,$ is congruent to a power of $,2,$ modulo $,p.,$ Note that the set $S$ of moduli where this fails is closed under multiplication. $\ \ $ – Bill Dubuque May 14 '24 at 23:51
  • Trivially, there are infinitely many primes $p_i$ larger than any $a=2^n$ for which $a\equiv 2^n \bmod p_i$. Perhaps the statement of your query can be tightened up to make clear that you wish to exclude this trivial case. – Keith Backman May 15 '24 at 16:48

1 Answers1

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Any positive integer $a$ works (e.g., $a = 3$, as suggested in Peter's comment). Consider

$$a = 2^{n}a_1, \;\; n \ge 0$$

where $a_1$ is a positive odd integer. For any odd prime $p_i$, since $2^n$ has a multiplicative inverse, we get

$$a \equiv 2^{n_i} \pmod{p_i} \, \iff \, a_1 \equiv 2^{n_i - n} \pmod{p_i}$$

Thus, we only need to only consider all odd integers $a_1$. As $a_1 = 1$ works (with $n_i = 0$ for all primes $p_i$), have $a_1 \gt 1$. Assume there are only a finite number of odd primes $p_i$, say $m$ of them, where there exists an integer $n_i$ with

$$a_1 \equiv 2^{n_i}\pmod{p_i}$$

Using the $p$-adic valuation function, among all of the odd prime factors (if any) $p_j$ of $a_1 - 1$ (since $n_i = 0$ shows these primes are a subset of the finite $p_i$ ones), let $k_1 = 1$ if there are no such primes, else have

$$k_1 = \max(\operatorname{ord}_{p_j}(a_1 - 1))$$

For an integer $k \gt k_1$ and using Euler's totient function, let

$$s = \prod_{i=1}^{m}\varphi(p_i^k)$$

Then, for each $1 \le i \le m$, we have

$$2^{s} - a_1 \equiv 1 - a_1 \not\equiv 0 \pmod{p_i^k} \;\;\to\;\; a_1 \not\equiv 2^{s} \pmod{p_i^k}$$

Thus, since $2^{s} - a_1$ can only contain at most $k_1$ of each of the prime factors among $p_i$, but $2^{s} - a_1$ is unbounded as $k$ grows, then it must have prime factors other any $p_i$ for sufficiently large $k$. However, this contradicts there are only a finite number of such primes, so there must be an infinite # of them.

John Omielan
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