x^2+y^2+z^2=xyz-1
Any help anyone? Thanks before hand.
Solution to what? A problem has a solution, not an equation! I suspect that this is the same problem that was posted on another board- in which it was specified that the problem was to find integer values of x, y, and z that satisfy this equation.
Let us assume that you want to find integers $\displaystyle x,y,z$ that satisfy the equation.
A check on the parity of each side of the equation shows that if the equation is to work, exactly two of $\displaystyle x,y,z$ must be even. Suppose WLOG that $\displaystyle x$ is odd and $\displaystyle y,z$ are even. Then $\displaystyle x^2\equiv1\mod 4$ (the square of any odd integer always $\displaystyle \equiv1\mod4)$ and $\displaystyle y^2 \equiv z^2 \equiv 0\mod4$ (the square of any even integer is divisible by $\displaystyle 4).$ Hence the LHS $\displaystyle \equiv1\mod4.$ On the other hand, $\displaystyle xyz\equiv0\mod4$ and so the RHS $\displaystyle \equiv-1\mod4.$ Hence the LHS can never equal the RHS, so there are no integer solutions to your equation.