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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 answer below).

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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. –  The Masked Avenger Apr 12 at 5:19
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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. –  The Masked Avenger Apr 12 at 5:26
    
@TMA, if you posted your question about $f(p)$, I would welcome it (while "similar interesting" seems vague and too encompassing). –  Włodzimierz Holsztyński Apr 12 at 19:07

3 Answers 3

up vote 7 down vote accepted

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.]

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Very good, thank you. –  Włodzimierz Holsztyński Apr 12 at 11:26

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 $7$.

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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). –  Włodzimierz Holsztyński Apr 12 at 12:20
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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? –  user46855 Apr 12 at 12:49
    
You are right. I could have a better look at my question before posting it on MO. I still like my question. –  Włodzimierz Holsztyński Apr 12 at 17:31
    
@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. –  user46855 Apr 13 at 6:56

There is also $(d p)=(2 3) $ :)

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Thank you. I will add your solution (see above, please). –  Włodzimierz Holsztyński Apr 12 at 7:47

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