<s>[INST] Modular calculation high exponent? I want to show, that $5^{96}\equiv -1 \pmod{193}$, without using the formula for quadratic residue.
So far I have : 
$5^{96}\equiv 5^{4\cdot24} \equiv 625^{24}\equiv 46^{24}\equiv 186^{12}\equiv -7^{12}\equiv 7^{12}\equiv 7^{3\cdot4}\equiv 150^4\equiv -43^4\equiv 43^4\equiv 112^2\equiv -81^2\equiv 3^8\\\
$
I think Euler's totient doesn't help, as $\varphi(193)=192>96=\frac{192}{2}$ or can I write this ?
$5^{96}\equiv 5^{192-96} \equiv 5^{-96}\equiv(5^{-1})^{96}\equiv 116^{96}\equiv -77^{96} \pmod{193} $
What am I doing wrong ? What am I missing ? Thanks in advance.
 [/INST] Let $x=5^{96}$. Then in the field $\Bbb F_{193}$ we have
$$
1=5^{\phi(193)}=(5^{96})^2=x^2,
$$
so that $(x-1)(x+1)=0$ in $\Bbb F_{193}$. Since a field has no zero divisors, we must have either $x=1$ or $x=-1$.
But $5^4=625=46$, so that $5^8=46^2=186$ and $5^{16}=49$,
$5^{32}=85$, so that
$$
5^{96}=(5^{32})^3=85^3=-1.
$$
Your calculation is correct, too, since
$$
3^8=6561=-1.
$$
</s>