Proving congruent statement for a prime number $p>10$

Question

I have the next statement that has to be proved:

Let $p$ be a prime number where $p>10$. Prove that $p-2$ has an inverse module $p$, this is, a number $q$ exists where $(p-2)q \equiv 1\mod(p)$.

I have this so far:

From $(p-2)q \equiv 1\mod(p)$ I know that $(p-2)q=kp+1$ and $p|((p-2)q-1)$.

So, $p|((p-2)q-1)$ is $p|pq-2q-1$.

I don't know which the next step should be exactly. Perhaps something like: $q=24k+r$ where $0 \le r<24$ and trying the different values $r$ can take?

Answer 1

Let $p=2k+1$. Then

$$2k+1 \equiv 0 \pmod{p}$$ $$2k \equiv -1 \pmod{p}$$ $$-2k \equiv 1 \pmod p$$ $$pk-2k \equiv 1 \pmod{p}$$ $$(p-2) k \equiv 1 \pmod{p}$$

Note that we actually proved that the reciprocal is $\frac{p-1}{2}$.

Answer 2

We want to show that there exist an inverse of $(p-2)$ modulo $p$, which is always true because since $p > 10$, $\gcd(p-2, p) = 1$.