Is there any element of order $51$ in the group $U(103)$ 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  $a$  

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

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