This is a proof proving that for a polynomial $f$ of degree $n$, $f(x)\equiv 0\pmod{p}$ has at most $n$ solutions. Is it correct?
For $n=1$, $$a_1x+a_0\equiv 0\pmod{p}$$ has exactly one solution since $\gcd(p,a_1)=1$. Suppose the proposition holds for $n-1$, that is, any polynomial of degree $n-1$ has at most $n-1$ solutions. Let $$f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_0.$$ If the congruence $f(x)\equiv 0\pmod{p}$ has no solutions, then the statement is true. If it has at least one solution, let $s$ be said solution. Observe that $$f(x)\equiv (x-s)g(x)\pmod{p}\quad g(x)\text{ has degree }n-1.$$ Since $g(x)$ has at most $n-1$ solutions and $s$ is one solution, the max number of solutions in the congruence is $n-1+1=n.\quad\square$