Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$

Question

Find all pairs $(x, y)$ of integers, such that $x^6 + 3x^3 + 1 = y^4$.

My solution:

Claim: The pair $(0, 1)$ are the only solutions.

Proof. Suppose there exists other solutions for $y \gt 1$ and $x \gt 0$, then I shall show that such pairs are impossible if $x$ and $y$ are integers. Let us Factorize the given equation as follows.

$x^6 + 3x^3 = y^4 - 1$

$x^6 + 3x^3 = (y^2)^2 - 1^2$

$x^6 + 3x^3 =(y^2 - 1)(y + 1)(y^2 + 1)$

$x^3(x^3+ 3) = (y - 1)(y + 1)(y^2 + 1) . . . (*)$

Because the numbers $(y - 1), (y + 1)$, and $(y^2 + 1)$ are all distinct, it follows that their products can never be a cube, hence, from $(*)$, we obtain the system.

$(y - 1)(y + 1)(y^2 + 1) = x^3 + 3$, and $x^3 = 1$, Which is equivalent to:

$(y - 1)(y + 1)(y^2 +) = 4 . . .(**)$

Since $(y - 1) \gt 0$, then $y$ is minimum if and only if $y = 2$ which clearly does not satisfy $(**)$. Hence we must have $x = 0$ and $y = 1$.

Please, Is there any mistake in my solution?.

Answer 1

$x^6+3x^3+1-y^4=0$ is a quadratic equation.

Thus, there is an integer $n$ for which $$3^2-4(1-y^4)=n^2$$ or $$n^2-4y^4=5$$ or $$(n-2y^2)(n+2y^2)=5$$ and we have four cases only.

Answer 2

Case 1. If $x= 0$ we have $y^4=1$ thus $y=\pm 1$

Case 2. If $x>0$ then

$(x^3+1)^2= x^6+2x^3+1< x^6 + 3x^3 + 1 = y^4$

$ y^4 = x^6 + 3x^3 + 1< x^6+4x^3+4 = (x^3+2)^2$

So we have:$$(x^3+1)^2< y^4 <(x^3+2)^2$$

So $x^3+1< y^2 < x^3+2$ a contradiction.

Case 3. If $x<0$ then write $x=-t$ and $t>0$. Now we have to sove:

$$t^6 - 3t^3 + 1 = y^4$$ and this can be done in similar fashion:

$$(t^3-2)^2<y^4<(t^3-1)^2$$