Diophantine equation no integer solutions

Question

Show that the following equation has no integer solutions:

$x^3+3x^2+2x=z^3-4z+4.$

No idea where to start because it has no $y$ functions.

Also I need to find the integer solutions to $y^2+x^2=9-z^2$.

Answer 1

Hint: Look at the equation mod $3$ using Fermat's Little Theorem.

More hints: $$ \begin{align} x^3&\equiv x&\pmod3\\ 3x^2&\equiv0&\pmod3\\ 2x&\equiv-x&\pmod3\\ \\ z^3&\equiv z&\pmod3\\ 4z&\equiv z&\pmod3\\ 4&\equiv1&\pmod3 \end{align} $$

Answer 2

$x^3+3x^2+2x=x(x+1)(x+2)$ divizible by $3$. Then applying mod 3 we get: $0=1$ because of $a^p = a \mod p$ when $p$ prime.