TITLE: Is there any element of order $51$ in the group $U(103)$
QUESTION [5 upvotes]: 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  $a$

REPLY [3 votes]: $103$ 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 $3 \cdot 17=51$.