algebra solving equation problem (10th grade standard) Solve this equation in the set of whole numbers:
$$2x^{3}+x+8=y^2$$ Please help!
 A: 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.
A: COMMENT.-Your equation $y^2=2x^3+x+8$ is one of elliptic curves for which the determination of a rational or integer points can be very easy as, for example, with $y^2=2x^3+x+7$ with apparent solution $(2,5)$ or can be very hard or even impossible with the known arithmetic tools.
When a solution is not easily visible, one can look for a convenient module that proves the impossibility when there is not solution but this can be difficult to find one. The correct and ingenious method  used by Peter  deserves your approval and even a brand best answer.
Peter verifies that the part $(2x^3+x)$ of the RHS is always congruent to $0$ modulo $3$ hence $$y^2=(2x^3+x)+8=(2x^3+x)+6+2\equiv 2\pmod 3$$ which is impossible. This is a correct solution.
A: I love Peter's answer, but I do not know many 10th graders that can come up with such a strategy. Another strategy would be to graph the two functions. Let $f(x) = 2x^3 + x + 8$ and let $g(x)=x^2$ and find the point(s) of intersection (where $f(x)=g(x)$).  A modern graphing utility shows a single point of intersection at roughly (-1.3, 1.8). This is clearly not an integer solution. 
However, the question asked for solutions in the set of whole numbers (positive integers). We note that when $x=0$, $f(0) = 8$ and $g(0) = 0$. In other words, $f(x) > g(x)$. As $x$ increases, $f(x)$ increases more rapidly than $g(x)$, maintaining the inequality. Thus $f(x) > g(x)$ for all $x > 0$, meaning the two functions will never intersect.
