No natural numbers satisfy $n\equiv n^2-4\pmod9$ Prove that for all natural numbers, $n$ is not congruent to $n^2-4\pmod9$.
I'm trying to prove this by contradiction and say that if they were congruent $\pmod 9$ it could be expressed as $9\mid n-(n^2-4)$ by the Modular Arithmetic theorem and then perhaps show that this cannot be expressed as $9k$ for some integer $k$ but I can't factor $n-(n^2-4)$ so I don't know where to go with this. Help appreciated
 A: You have to consider $n^2  - 4 - n \mod 9$. So this is the same as  $n^2 + 8 n - 4 \mod 9$. And this is $(n+4)^2 - 20 \mod 9$ or $(n+4)^2 - 2 \mod 9$. 
Now just check that $2$ is not a square modulo $9$. 
A: When 
$$n\equiv 0,1,2,3,4,5,6,7,8\pmod 9$$
one has
$$n^2-4\equiv 5,6,0,5,3,3,5,0,4\pmod 9$$
respectively, and one can see that $n$ is never congruent to $n^2-4$ 
A: Consider the fact that $n \equiv 0,\pm1,\pm2,\pm3,\pm4 \pmod{9}$. Then $n^2 \equiv 0,1,4,0,7 \pmod{9}$. This leads to $n^2-4 \equiv 5,6,0,5,3 \pmod{9} \,$ respectively, which is not the same as $n \equiv 0,\pm1,\pm2,\pm3,\pm4 \pmod{9} \,$ respectively.
A: Write 
$$4n^2-4n-7\equiv 0\pmod 9$$
or
$$(2n-1)^2+1\equiv 0\pmod 9$$
Since $-1$ is not a square mod $3$, this has no solutions.
A: If $n \equiv n^2 - 4 (\mod 9)$ then $3 | n^2 -n -4$, so $n^2 -n -1 \equiv 0 (\mod 3)$. But, in $\mathbb{Z}_3$, none of $\hat{0}, \hat{1}, \hat{2}$ verifies this equation, so the original relationship cannot hold.
A: It is the special case $\ p = 3\ $ below (since a root $\bmod 9\,$ persists $\!\bmod 3)$
Theorem $\bmod p\!:\,\ f(x)\, =\, x^{p-1}-x-1\,$ has no roots.
Proof $\ $ By $\mu$Fermat $\,a\not\equiv 0\,\Rightarrow\, f(a) \equiv -a\not\equiv 0,\, $ and $\,f(0)\equiv -1\not\equiv 0$
