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 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.   
