Are These the Only Integer Solutions to $y^2 = x^4 + x^3 + x^2 + x + 1$?

In summary, the equation $y^2=x^4+x^3+x^2+x+1$ only has integer solutions at $(-1,\,\pm 1)$, $(0,\,\pm 1)$, and $(3,\,\pm 11)$. This can be proven using Descartes' rule of signs, which states that a polynomial equation can have at most as many positive roots as there are sign changes in the coefficients and at most as many negative roots as there are sign changes in the coefficients of the terms with even powers. Since there are no sign changes in the coefficients of the given equation, it follows that there can only be a maximum of 2 positive roots and 2 negative roots. By inspection
  • #1
anemone
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Prove that $(-1,\,\pm 1)$, $(0,\,\pm 1)$, $(3,\,\pm 11)$ are the only integers solution for the equation $y^2=x^4+x^3+x^2+x+1$.
 
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  • #2
$x,y $ are integers
let :$y^2=x^4+x^3+x^2+x+1=k-----(1)$
here $k\in N$ ,and $k$ is a perfect square
we have :$(x-1)(x^4+x^3+x^2+x+1)=k(x-1)$
or $x^5-kx+k-1=0----(2)$
if $k=1$ then the solutions of (1):$(x,k)=(-1,1),(0,1),$and two complex conjugates
soluions of (2):$(x,k)=(-1,1),(0,1),(1,1) $ and two complex conjugates
if $k>1$ then the solutions of(1):1 positive , 1 negative and two complex conjugates(idependent of k)
the solutions of(2):2 positive , 1 negative and two complex conjugates
we are given :the solutins of (1):$(x,k)=(-1,1),(0,1),(3,121)$
and we have the soltions of (2):$(x,k)=(-1,1),(0,1),(3,121),(1,k)$
so $(x,y)=(-1,\pm 1),(0,\pm1),(3,\pm 11) $ are the only possible solutions
 
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  • #3
Hey Albert, I'm sorry because I can't see your point here, about how you deduced the number of roots and how they are independent of $k$...but I still want to thank you for attempting.

Without any further delay, I will share the solution that I have at hand here, I hope you and others will like it and perhaps, you want to explain to us more about your solution?

Suppose $(x,\,y)$ is a point with integer coordinates on the given curve.

From the facts that

$\left(x^2+\dfrac{x}{2}\right)^2=x^4+x^3+\dfrac{x^2}{4}=y^2-\dfrac{3x^2}{4}-x-1=y^2-\dfrac{3}{4}\left(x+\dfrac{2}{3}\right)^2-\dfrac{2}{3}<y^2$ and

$\left(x^2+\dfrac{x}{2}+1\right)^2=x^4+x^3+\dfrac{9x^2}{4}+x+1=y^2+\dfrac{5x^2}{4}\ge y$, we conclude that $x^2+\dfrac{x}{2}<|y| \le x^2+\dfrac{x}{2}+1$.

If $x$ is odd, then $|y|=x^2+\dfrac{x+1}{2}$ is the only integer in this interval and

$\begin{align*}y^2&=\left(x^2+\dfrac{x+1}{2}\right)^2\\&=x^4+x^3+x^2+\dfrac{x^2+2x+1}{4}\\&=x^4+x^3+x^2+x+1+\dfrac{x^2-2x-3}{4}\\&=y^2+\dfrac{(x-3)(x+1)}{4}\end{align*}$

It follows that $(x-3)(x+1)=0$ and so $x=3$ or $x=1$. This gives us the integer points $(3,\,\pm 11)$ and $(-1,\,\pm 1)$.

If $x$ is even, then $x^2+\dfrac{x}{2}<|y| \le x^2+\dfrac{x}{2}+1$ implies that $|y|=x^2+\dfrac{x}{2}+1$. Then $\left(x^2+\dfrac{x}{2}+1\right)^2=x^4+x^3+\dfrac{9x^2}{4}+x+1=y^2+\dfrac{5x^2}{4}\ge y$ implies that $y^2=y^2+\dfrac{5x^2}{4}$ and therefore $x=0$, giving us the integer points $(0,\,\pm 1)$ and so these are the only solutions.
 
  • #4
Albert said:
$x,y $ are integers
let :$y^2=x^4+x^3+x^2+x+1=k-----(1)$
here $k\in N$ ,and $k$ is a perfect square
we have :$(x-1)(x^4+x^3+x^2+x+1)=k(x-1)$
or $x^5-kx+k-1=0----(2)$
if $k=1$ then the solutions of (1):$(x,k)=(-1,1),(0,1),$and two complex conjugates
soluions of (2):$(x,k)=(-1,1),(0,1),(1,1) $ and two complex conjugates
if $k>1$ then the solutions of(1):1 positive , 1 negative and two complex conjugates(idependent of k)
the solutions of(2):2 positive , 1 negative and two complex conjugates
we are given :the solutins of (1):$(x,k)=(-1,1),(0,1),(3,121)$
and we have the soltions of (2):$(x,k)=(-1,1),(0,1),(3,121),(1,k)$
so $(x,y)=(-1,\pm 1),(0,\pm1),(3,\pm 11) $ are the only possible solutions
explanation here:
Descartes' rule of signs - Wikipedia, the free encyclopedia
 
  • #5


I would approach this problem by first analyzing the given equation. The equation $y^2=x^4+x^3+x^2+x+1$ is a polynomial equation of degree four, which means it can have a maximum of four solutions. Additionally, since the equation only contains terms with even exponents, we can assume that both $x$ and $y$ are even or both are odd.

Next, I would consider the given solutions $(-1,\,\pm 1)$, $(0,\,\pm 1)$, $(3,\,\pm 11)$ and see if they satisfy the equation. Upon substitution, I can verify that all six of these solutions indeed satisfy the equation.

Now, to prove that these are the only integer solutions, I would use a proof by contradiction. I would assume that there exists another solution $(a,b)$ where $a$ and $b$ are integers. This means that $a^4+a^3+a^2+a+1=b^2$. Rearranging this equation, we get $b^2-a^4=a^3+a^2+a+1$. This can be factored as $(b-a^2)(b+a^2)=a(a^2+a+1)+1$.

Since both $b$ and $a$ are integers, $b-a^2$ and $b+a^2$ must also be integers. This means that $a(a^2+a+1)+1$ must be divisible by both $b-a^2$ and $b+a^2$. However, the only possible integer factors of $a(a^2+a+1)+1$ are $\pm 1$ and $\pm (a^2+a+1)$, as all other factors would make the expression too large. This means that either $b-a^2=\pm 1$ or $b+a^2=\pm 1$.

If $b-a^2=\pm 1$, then $b=a^2\pm 1$. Substituting this into the original equation, we get $a^4+a^3+a^2+a+1=(a^2\pm 1)^2$, which simplifies to $a^2(a^2\pm 2)=0$. This means that $a=0$ or $a=\pm 1$, which are already included in the given solutions.

If
 

FAQ: Are These the Only Integer Solutions to $y^2 = x^4 + x^3 + x^2 + x + 1$?

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