4

What are all integral solutions to $$y^2=x^4+x^3+x^2+x+1$$

The RHS could become $$(x^2-x+1)(x^2+x+1)+x(x^2+1)$$ or $$\frac{x^5-1}{x-1}$$ I have no idea how to manipulate the equation into something useful or what the first step should be. Also, A quartic diophantine equation looks useful, but none of the answers completely solve the question? Thanks!

MJD
  • 67,568
  • 43
  • 308
  • 617
Baker5680
  • 451

2 Answers2

4

If $x \gt 0$, the only solution is $x=3, y=11$.

Note: Wolfy was used extensively getting this answer.

$y^2 =x^4+x^3+x^2+x+1 $

$y^2 =x^4+x^3+x^2+x+1 \gt x^4 \implies y > x^2 $.

$(x+1)^4 =x^4+4x^4+,,, $ so $y^2 \lt (x+1)^4 $ or $y < (x+1)^2 $.

$(x^2+x/2+3/8)^2 =x^4 + x^3 + x^2 + (3 x)/8 + 9/64 \lt y^2 $ so $y > x^2+x/2+3/8$.

$(x^2+x/2+1)^2 =x^4 + x^3 + (9 x^2)/4 + x + 1 \gt y^2 $ so $y < x^2+x/2+1 $.

If $x = 2n$ then $4n^2+n+3/8 \lt y \lt 4n^2+n+1 $, so there can be no such integer $y$.

If $x = 2n+1$ then $x^2+x/2+3/8 =4n^2+4n+1+n+1/2+3/8 =4n^2+5n+15/8 $ and $x^2+x/2+1 =4n^2+4n+1+n+1/2+1 =4n^2+5n+5/2 $ so $y = 4n^2+5n+2 $.

But $y^2 =16 n^4 + 40 n^3 + 41 n^2 + 20 n + 4 $ and $x^4+x^3+x^2+x+1 =\dfrac{x^5-1}{x-1} =\dfrac{(2n+1)^5-1}{2n} =16 n^4 + 40 n^3 + 40 n^2 + 20 n + 5 $ and the difference is $n^2-1$ so they are never equal unless $n = 1$ so $x = 3 $.

marty cohen
  • 110,450
  • 1
    A minor point, but I think your start of "If $x \gt 0$, ..." should be something like "If $x,y \gt 0$, ..." instead since otherwise, for example, $x = 3, y = -11$ is also a solution. – John Omielan Oct 22 '19 at 01:56
  • 1
    Very good solution! I believe $n=-1$ yields another case, right? – Baker013273213 Oct 22 '19 at 16:52
  • 2
    @Baker013273213 Change $x$ to $-z$ and assume $z>0$. $z$ cannot be even, otherwise $$(z^2-z/2)^2 < z^4-z^3+z^2-z+1=y^2 < (z^2-z/2+1)^2$$ With $z$ odd, we have $$(z^2 - (z+1)/2)^2 < z^4-z^3+z^2-z+1=y^2 < (z^2 - (z+1)/2+1)^2$$ if $z > 1$. So only possibility is $z=1$ giving $x=-1$. – Yong Hao Ng Oct 23 '19 at 02:59
  • 1
    You should make this an answer. – marty cohen Oct 23 '19 at 14:35
2

There is a result of Masser [1] that $$y^2 = f(x) \tag 1$$ where $f(x)$ is a monic polynomial and not a perfect square, any integer solution $(x,y)$ of Eqn. (1) satisfies the inequality

$$ |x| \le 26 H(f)^3 \tag 2 $$

where $H(f)$ denotes the classical height of $f$ given by the maximum of the absolute values of the coefficients of $f(x)$.

In the given equation, $H(f) = 1$. Therefore, all integer solutions are given by $x \in [-26, +26]$.

This range is quite small and effective for a computer search.

A brute force search in this range yields

$$(x,y) \in \{(-1,1), (0,1), (3,11)\}, \text{ where } (x,y) \in \mathbb{Z\times Z}.$$

Note: Checking for small root $x \in \{0,\pm 1, \pm 2, \cdots \}$ and eliminating a factor to reduce to a cubic or quadratic is the normal method followed. We could have used that as suggested in other answers and comments. The method above is useful when such an elimination is not possible or tedious. Therefore, I am providing this method for reference of other readers who may find this numerical method useful, in general.

References

[1]: Masser, D. W. “Polynomial Bounds for Diophantine Equations.” The American Mathematical Monthly, vol. 93, no. 6, 1986, pp. 486–88. JSTOR, https://doi.org/10.2307/2323487. Accessed 6 Dec. 2022.

vvg
  • 3,526