Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$

Question

I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.

I have a couple of ideas for going about this exercise.

$1)$ By moving $1$ to the other side of the equation we obtain:

$y^4-1=x^4-x^3+x^2-x \rightarrow (y^2-1)(y^2+1)=x(x-1)(x^2+1)$.

The LHS gives two consecutive integers I am able to see, but other than that I am stuck and not sure where to go from here.

$2)$ We notice that $1-x+x^2-x^3+x^4=\Phi_{10}(x)=y^4$.

We now have that $\Phi_{10}(x) \geq 0$ since $y=\pm \sqrt[4]{1-x+x^2-x^3+x^4}$.

In this case, with some computation, $x={0,1}$ are the only possibilities that will force $y \in \mathbb{Z}$. From there, I use induction to show that $x \geq 2$ and $x \leq -1$ do not yield a perfect fourth.

In either case, I keep running into some issues. If someone could please other a hint to help me continue through this exercise, that would be very helpful. Thank you.

Update:

Hello, my progress on the following problem is as follows:

Suppose that $x=y$, then we get $y^3-y^2+y-1=0 \rightarrow y=1, \pm i$. Thus we get two solutions $(x,y)={(1,1),(1,-1)}$, the $y=-1$ since $y^4$ was in the original equation.

Next, using @Batominovski hint we can write:

$(2x^2-x)^2 < (x^4-x^3+x^2-x+1) < 4(x^4-x^3+x^2-x+1) \leq (2x^2-x+2)^2 \forall x \in \mathbb{R}$ which is true for $x=0$.

Since the multiplication of a constant (in this case $k=4$) does not affect a solution from existing, we have that $x=0$ is a solution, consequently we find two more solutions are $(x,y)=(0,1),(0,-1)$.

Answer 1

Hint: Note that $$\left(2x^2-x\right)^2<4\left(x^4-x^3+x^2-x+1\right)\leq\left(2x^2-x+2\right)^2$$ for all $x\in\mathbb{R}$. The right-hand side becomes an equality if and only if $x=0$.

Answer 2

Hint: note that for $x$ big enough the following holds: $$(x-1)^4 < x^4-x^3+x^2-x+1 < x^4.$$ Similarly, for $x$ small enough we have $$(x-1)^4 > x^4-x^3+x^2-x+1 > x^4.$$ Therefore you are left with a finite number of possible values of $x$.

Answer 3

One has $$1-x+x^{2}-x^{3}+x^{4}=y^{4}\iff(2x^2+1)^2=4y^4+4(x^3+x)-3$$ Notice that necessarily $y$ must be odd and $x^3+x$ is even when $x$ is even or odd. Hence one has $$(2x^2+1)^2\equiv 4y^4-3 \pmod 8$$ The solutions of this congruence are $$(x,y)=(x,1),(x,3),(x,5),(x,7)\text { with } x=0,1,2,3,4,5,6,7$$ we can verify that $\color{red}{(x,y)=(0,\pm1),(1\pm 1)}$ are the only solutions because for $y=1+8m,3+8m,5+8m,7+8m$ we have for $m=0$ these only solutions (easier to verify for other candidates with $m\gt 1$making $1-x+x^{2}-x^{3}+x^{4}=\frac{x^5+1}{x+1}$)