algebra solving equation problem (10th grade standard) Solve this equation in the set of whole numbers:
$$2x^{3}+x+8=y^2$$ Please help!

Consider $$2x^3+x=x(2x^2+1)$$
This is always divisible by $3$ because in the case $x=3k+1$, we have $2x^2+1=18k^2+12k+3$ and in the case $x=3k+2$ , we have $2x^2+1=18k^2+24k+9$. 
Therefore, the left side has the form $3k+2$ for every ineteger $x$, but there is no square of the form $3k+2$ because the possible remainders modulo $3$ are $0$ and $1$.
Hence, the equation is not solveable over the integers.
