Integer solutions Challenge

[anemone]

Prove that $(-1,\,\pm 1)$, $(0,\,\pm 1)$, $(3,\,\pm 11)$ are the only integers solution for the equation $y^2=x^4+x^3+x^2+x+1$.

 

[Albert1]

Spoiler

$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

 

[anemone]

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?

Spoiler

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.

 

[Albert1]

Spoiler

$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

 

[CellCoree]

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