primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$

The number $32$ is small, so it is easy to show by computing that $5$ has order $8$. Note that we only need to compute $5^{2^k}$ modulo $32$. For $\varphi(32)=16$, so the order of $5$ must be a power of $2$. 
The following is severe overkill. We show that for any $n\ge 3$, $\mathbb{Z}^\ast_{2^n}$ is generated by $5$ and $-1$.
Lemma: Let $n\ge 3$. Then $5^{2^{n-3}}\not\equiv 1 \pmod{2^n}$, while $5^{2^{n-2}}\equiv 1\pmod{2^n}$.
Proof of Lemma: We show by induction on $n$ that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. This easily yields both parts of our assertion.
The result is easy to check for $n=3$.   Suppose now that 
$$5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^{k}}.\tag{$1$}$$ We show that $5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$. 
 Square both sides of $(1)$, and simplify modulo $2^{k+1}$. We get
 $$5^{2^{k-2}}=(1+2^{k-1} +n2^k)^2\equiv 1+2^{k}\pmod{2^{k+1}},$$
 since $2^{2k-2}$ is divisible by $2^{k+1}$ if $k\ge 3$.  This ends the proof of the lemma.
Now it is easy to finish.  The order of $5$ modulo $2^n$ divides $\varphi(2^n)=2^{n-1}$, so the order of $5$ is a power of $2$. It follows from the results above that the order of $5$ is exactly $2^{n-2}$.
No power of $5$ is congruent to $-1$ modulo $2^n$. It follows that if $n \ge 3$, then $5$ and $-1$ together generate all of $\mathbb{Z}^\ast_{2^n}$. 
