# Thread: Polynomial, area of a polygon

1. ## Polynomial, area of a polygon

Hi!

Could you help me solve this problem?

Integers $(x,y)$ satisfying the equation
$xy^2 - y^3 - xy + x^2 + 5 =0$

are vertexes of a certain polygon. What is its area?

Thank you.

2. ## Re: Polynomial, area of a polygon

Hello, gollum!

I used some Olympic-level gymnastics . . .

Integers $(x,y)$ satisfying the equation: $xy^2 - y^3 - xy + x^2 + 5 \:=\:0$

are vertexes of a certain polygon. .What is its area?

We have a quadratic in $x:\;\;x^2 - (y-y^2)x - (y^3-5) \:=\:0$

Quadratic Formula: . $x\;=\;\frac{(y-y^2) \pm \sqrt{(y-y^2)^2 + 4(y^3-5)}}{2}$

. . $x \;=\;\frac{(y-y^2) \pm\sqrt{y^4 + 2y^3 + y^2 - 20}}{2} \;=\;\frac{(y-y^2) \pm \sqrt{y^2(y+1)^2 - 20}}{2}$

Since $x$ is an integer, the radicand must be a square.
. . . . . . $y^2(y+1)^2 - 20 \:=\:k^2$

Hence, we have: . $\big[y(y+1)\big]^2 - k^2 \:=\:20$

The difference of two squares is 20.
. . The only such squares are: . $(\pm6)^2\text{ and }(\pm4)^2$

This gives us: . $y = 2,\;\begin{Bmatrix}x = 1 \\ x=\text{-}3\end{Bmatrix}$

. . . . . . and: . $y = \text{-}3,\;\begin{Bmatrix}x = \text{-}4 \\ x = \text{-}8\end{Bmatrix}$

We have four vertices:

. . $\begin{array}{ccccccc} && * &&& * \\ && (\text{-}3,2) &&& (1,2) \\ \\ \\ * &&& * \\ (\text{-}8,\text{-}3) &&& (\text{-}4,\text{-}3) \end{array}$

We have a parallelogram with base 4 and height 5 . . .