Find all the solutions of the polynomial equation $x^5+y^3 \equiv 1 \pmod9$. Find all the solutions of the polynomial equation $$x^5+y^3 \equiv 1\pmod9.$$ I have learned to find the solutions of the polynomials like $x^p \equiv n\pmod m$, but never been like two variables. Please help me to solve this. Thanks.
 A: Guide:
Let $y=3m+r, r=0,1,2$
$y^3=(3m+r)^3=(3m)^3+3(3m)^2r+3(3m)r^2+r^3\equiv r^3 \pmod{9}$
Hence we just have to consider $3$ cases, 
Case $1$: $y \equiv 0 \pmod{3}$ and the problem reduces to $x^5 \equiv 1\pmod{9}$.
Case $2$: $y \equiv 1 \pmod{3}$ and the problem reduces to $x^5 \equiv 0\pmod{9}$.
Case $3$: $y \equiv -1 \pmod{3}$ and the problem reduces to $x^5 \equiv 2\pmod{9}$.
The problems are now of the form of $x^p \equiv n \pmod{m}$
A: We need to find $x$ and $y$ for which $$ x^5+y^3 \equiv 1\pmod9$$
We check the remainders of $x^5$ and $y^3$ in dividing by 9 for possible solutions.
Solutions mod 9 are:
$$x=1, y\in \{3,6,0\}$$  or $$ x\in \{3,6,0\}\text {and }  y\in \{1,4\}$$
A: Siong’s method is much more elegant, but as Jyrki said, this problem can be brute forced quite easily:
Step 1: Calculate all possible integer cubes $a^3$, where $a\in \mathbb{Z}_9 $ (for instance $4^3=64 \equiv 1 \pmod 9$).
Step 2: Repeat Step 1, but this time for all possible values of $a^5$ (this is the most tedious step). 
Step 3: Simply check which solutions are possible by simple arithmetic (thus, for example, by above we already know that when $y=4$, $x=0,3,6$ are all solutions).
