# Diophantine equation no integer solutions

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 $y^2+x^2=9-z^2$.

• For your second equation note that $x^2+y^2+z^2=9$, so $-3\le x,y,z\le 3$. – Xam Dec 1 '16 at 15:22
• For the second question, you should really make a second post. However, there are only a few possibilities for $x^2+y^2+z^2=9$ since $|x|,|y|,|z|\le3$. – robjohn Dec 1 '16 at 15:22
• If you change all the $z$s to $y$ in the first equation then you have $y$ functions but no $z$ functions. Is that better? – David K Dec 1 '16 at 15:22
• If i change z to y, factorise both side to x(x+1)(x+2)-y(y-2)(y+2)=z^2 , where z=2. Where do i go from there, or am i going in the wrong direction? – K_uddin Dec 1 '16 at 15:44

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