0
$\begingroup$

Solve in integers $y^3-1=x^4+x^2$ First you can add $1$ and factor to get $y^3=(x^2+x+1)(x^2-x+1)$ Since the GCD of $(x^2+x+1)$ and $(x^2-x+1)$ is $1$, then they both must be perfect cubes. I don't know what to do after that. Any help? Thanks

$\endgroup$
4
$\begingroup$

As you have noted, by rearranging the equation you can find that $y^3=(x^2+x+1)(x^2-x+1)$, and by considering the GCD of the two polynomials on the RHS you know that both must be perfect cubes.

Note that one solution is $(x,y)=(0,1)$.

By symmetry, we may assume that $x> 0$ for all other solutions.

Consider $(x^2+x+1)-(x^2-x+1)=2x<2(x^2+x+1)^{1/2}$. This says that the difference between our two cubes must be less than twice the square root of the larger cube. However, the difference between any two cubes $(n+k)^3-n^3>3n^2$ is larger than thrice the square of the cube root of the smaller cube.

Thus, $3(x^2-x+1)^{2/3}<2(x^2+x+1)^{1/2}$, which is a contradiction if $x>1$.

Therefore, the only possible solutions beside $(x,y)=(0,1)$ must have $x=\pm 1$, which obviously do not work.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.