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Does there exist an element of order $51$ in the multiplicative group $U(103)$ ?

Now if the element exist say $x$ then it satisfies the equation $$x^{51}\equiv 1\pmod {103}$$ . Now $103$ being a prime it is clear that there is an element $y$ in $U(n)$ satisfying $$y^{102}\equiv 1\pmod {103}$$ So exactly an element of half order is required to be found.

Is there any result in number theory that might imply that $$a^{p-1}\equiv 1\pmod p$$ ensures the existence of some $b$ such that $$b^{{p-1}\over {2}}\equiv {1}\pmod p$$ where $p$ is a prime not dividing $a$


marked as duplicate by Servaes, PhoemueX, user147263, egreg, N. F. Taussig Sep 24 '15 at 1:11

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  • 2
    $\begingroup$ What's the order of $y^{2}$? $\endgroup$ – Josh B. Aug 23 '15 at 22:20
  • $\begingroup$ that is $51$. Thank you. $\endgroup$ – user118494 Aug 23 '15 at 22:21
  • $\begingroup$ Take $b=a^2$, and $a$ a generator of $U(103)$ (don't worry about the existence of $a$). $\endgroup$ – Yassine Guerboussa Aug 25 '15 at 0:51
  • $\begingroup$ @Josh B : What is the order of $y^2$? $\endgroup$ – Yassine Guerboussa Aug 25 '15 at 0:59
  • $\begingroup$ @Servaes : In the link you posted , the user just asked a question . In my post I have , although the same problem, attempted something on my own and asked whether that method is correct and can be furthered to find the result . Right $?$ . Why flag it as duplicate $?$ $\endgroup$ – user118494 Sep 23 '15 at 12:22

$103$ is a prime and so $U(103)$ has $103-1=102 = 2 \cdot 3 \cdot 17$ elements.

By Cauchy's theorem, $U(103)$ has an element $a$ of order $3$ and an element $b$ of order $17$.

Since $U(103)$ is abelian, $ab$ has order $3 \cdot 17=51$.


I suppose that $U(103)$ denotes the multiplicative group of the ring $\mathbb{Z}_{103}$.

Note that $\mathbb{Z}_{103}$ is a field; the multiplicative group of a finite field is cyclic; and a cyclic group of order $n$ contains elements of order $d$ whenever $d$ divides $n$ (take the $\frac{n}{d}$-th power of a generator).

You can replace $103$ by some non-primes $n$, and still $U(n)$ remains cyclic.

  • $\begingroup$ If I replace 103 by 24 would $U(24)$ be cyclic ? @Yassine Guerboussa $\endgroup$ – user118494 Aug 25 '15 at 15:28
  • $\begingroup$ No, $U(24)$ is elementary $2$-abelian of rank $3$ (a vector space of dim $3$ over the field with two elements). In general, for an prime $p$, we have $U(p^n)$ is cyclic of order $(p-1)p^{n-1}$; and $U(2^n)$ is a product of the cyclic group of order $2$ by the cyclic group of order $2^{n-2}$ (for $n\geq 3$). You can find $U(n)$ easily from the decomposition of $n$ as a product of prime powers. $\endgroup$ – Yassine Guerboussa Aug 27 '15 at 20:39

I found that $2$ is an answer, but I just used mathematica, which might not be what you're looking for.

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Note: also found $4,6,7,15$. Probably more.

  • $\begingroup$ There are $32=\phi(51)$ solutions, but $11$ is not one of them. $\endgroup$ – lhf Aug 25 '15 at 1:34
  • $\begingroup$ It's been awhile since I've used Mathematica, so I might be slightly off, but I believe the following will tell you every answer: Select[Range[2..102], AND[Mod[#^3,103]!=1,Mod[#^17,103]!=1,Mod[#^51,103]==1]&];. $\endgroup$ – zibadawa timmy Aug 25 '15 at 1:49

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