1
$\begingroup$

NOTATION: $O_x$ -- the product of all odd primes $\le x$.

E.g. $O_7=3\cdot 5\cdot 7 = 105$.

QUESTION: Are the three ordered pairs $\ (d\ p)=(1\ 3)\ \ (2\ 3)\ \ (4\ 5)\ $ the only solutions of the equation: $$|O_p-2^d|=1$$ in natural numbers $d$, and odd primes $p$?

(I don't know an answer).

MOTIVATION: Let $\ s\ $ be a prime just after $\ p$, so that $\ p<s$. If $$|O_p-2^d|\ne 1$$ then $$|O_p-2^d|\ge s$$ Furthermore, sometimes $\ s\ $ can be quite a bit larger than $\ p$.

ACKNOWLEDGEMENT: Bjørn Kjos-Hanssen has provided one of the above solutions (see his comment below).

Improve this question
$\endgroup$
4
  • $\begingroup$ Likely yes. For large p Fermat's little theorem will determine the character of the primes dividing 2^p +- 1. You can check this out using Carmichael's tables to shoa no other small solutions. $\endgroup$
    – The Masked Avenger
    Commented Apr 12, 2014 at 5:19
  • 1
    $\begingroup$ You might prefer the following. Let f(p) be the power of 2 that divides precisely (O_p)^2 - 1. Is f(p) unbounded as a function of p? Similar interesting questions around f(p) could be posed. $\endgroup$
    – The Masked Avenger
    Commented Apr 12, 2014 at 5:26
  • $\begingroup$ @TMA, if you posted your question about $f(p)$, I would welcome it (while "similar interesting" seems vague and too encompassing). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Apr 12, 2014 at 19:07
  • $\begingroup$ There is also $(d p)=(2 3) $ :) $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Apr 2, 2023 at 5:56

3 Answers 3

Reset to default
9
$\begingroup$

There is no solution for $O_n=2^d-1$ with $n \geq 7$.

If 5 divides $2^d-1$ and 7 divides $2^d-1$, then 9 divides $2^d-1$. [Because 4 divides $d$ and 3 divides $d$; 6 divides $d$ and hence $2^{d}-1$ is divisible by 9.]

Improve this answer
$\endgroup$
0
3
$\begingroup$

Let's complete the answer following The Masked Avenger: modulo $7$ the powers of $2$ are $2,4,1$ and then repeat cyclically, so that $1+2^d$ is never divisible by $7$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you. Hm, this case was even simpler than $2^d-1$. Now, which answer should I award? I hope you will not be upset that I will choose the more involved case (even if both turned out to be easy). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Apr 12, 2014 at 12:20
  • 1
    $\begingroup$ I think nobody here should be upset. Let's see: five people for a question that a high school student could solve. Is this (math) over or under flow? $\endgroup$
    – user46855
    Commented Apr 12, 2014 at 12:49
  • $\begingroup$ You are right. I could have a better look at my question before posting it on MO. I still like my question. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Apr 12, 2014 at 17:31
  • 1
    $\begingroup$ @Wlodzimierz Holsztynski: my sincere apologies, only now I realize that my comment could be negatively taken. I am not qualified to judge what is suitable or not for MO (I only joked about the number of people, including myself, and time necessary to reach the conclusion "yes, it is obvious"). In my opinion neat and briefly & elementary solvable problems like yours deserve a wider audience than the MO one, a audience including students and teachers. Not being MO the best forum for your question is not a negative connotation. $\endgroup$
    – user46855
    Commented Apr 13, 2014 at 6:56
0
$\begingroup$

The only solutions to the Diophantine equation $O_x=2^y-1$ with $x$ prime are $(2,1)$, $(3,2)$ and $(5,4)$ (assuming $O_2=1$ is the empty product). Proof:

Suppose $x\ge7$. The number $O_x$ is divisible by $21$ but not by $63$ since $O_x$ is squarefree. Hence $2^y-1$ is divisible by $21$, but not by $63$, so $O(x)\equiv21,42 (\mod 63)$.

$2^x\mod 63$ is $1, 2, 4, 8, 16, 32$ for $x\mod6=0, 1, 2, 3, 4, 5$, respectively. Hence if $x$ is an integer, then $2^x-1\equiv0, 1, 3, 7, 15, 31$ mod 63.

This gives a contradiction, so $x\lt7$. The remaining cases can be done by hand.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .