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$32^n = 167x + 2$, find the smallest positive integer n for some integer x.

I have searched the mathematics exchange site for this question, but there might be some mistake in my searching capabilities please provide the link if the same question is found anywhere else

My thought process:
I thought of making x = 2y, (bcz LHS is even) then it converts to $32^n = 334y + 2$ then dividing by 2
$2^{5n-1} = 167y + 1$ then thought of factorisation, but wasnt able to proceed then againg tried mod 16 got that y = 16k + 9 format

After some of this I realised that I have to use bounding to get the upperbound of n (upperbound of x would also work the same) Now I am asking for help, as I am not able to proceed in the direction of bounding.
This question is from PUMaC 2008-09

Bill Dubuque
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  • key word: Discrete logarithm. Quote from Wikipedia "Discrete logarithms are quickly computable in a few special cases. However, no efficient method is known for computing them in general." . That is, unless there is some neat trick, it's not easy to find a solution, that's all I can tell... – Quý Nhân May 09 '25 at 06:06
  • We do get valid solutions if $5n-1$ is a multiple of $\phi(167)=166$ (by Euler's theorem), so $n=133$ works. But that doesn't confirm that it's the smallest possible value of $n$ – Soham Saha May 09 '25 at 06:21
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    Also, $2^{\frac{\phi(167)}2}$ turns out to be $1$ (mod $167$) So $n=50$ also works. – Soham Saha May 09 '25 at 06:30
  • This boils down to computing the order of $2$ modulo $167$ for which the common methods are already described here in many places, e.g. see the linked dupe. – Bill Dubuque May 24 '25 at 02:45

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The question is equivalent to finding the minimal $n$ such that $$2^{5n-1}\equiv 1\mod 167,$$ so it suffices to calculate the order $k$ of $2$ modulo $167$, and the desired $n$ is the minimal one such that $k|5n-1$.

By Fermat's little theorem, we have $2^{166}\equiv 1\pmod{167}$, hence $k|166=2\times83$. The only possible $k$'s are $1,2,83$ and $166$. Clearly $k\neq 1,2$, hence it suffice to show whether $2^{83}\equiv \pm1\pmod{167}$. This can be seen by the Euler's criterion:

$$\left(\frac{2}{167}\right)\equiv 2^{83}\pmod{167}.$$

We are lucky in this case: $13^2=169\equiv 2\pmod{167}$. Hence $\left(\frac{2}{167}\right)=1$. Therefore $k=83$.

To conclude, the minimal $n$ such that $83|5n-1$, is $\boxed{50}$.

Yunfeng Gong
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  • Please strive not to answer questions that are duplicates of FAQS, cf. site policy. Most basic question like this have already been answered many times over so will usually be closed as dupes. – Bill Dubuque May 24 '25 at 02:47
  • Dear @BillDubuque : The answer I provide here is not really similar to the answers of the duplicated question (which are by tedious and laborious computation). Computing 2^83 modulo 167 by hand is really not realistic. – Yunfeng Gong May 24 '25 at 09:41
  • Applying Euler's criterion / reciprocity is already mentioned here in the linked dupe (and many other places). The site is 15 years old. There is little novel that can be said on basic matters like these. $\ \ $ – Bill Dubuque May 24 '25 at 10:43