Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$

Question

Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$

My attempt :

$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$

$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$

$=(xy+(x^2-y^2))^3$

$=(x^2+xy-y^2)^3 = 1$

so $x^2+xy-y^2 = 1$

Please suggest how to proceed.

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Thank you, AmateurMathGuy and lhf.

Please check my work on Induction for Fibonacci sequence.

$$t_{k+2}-3t_{k+1}+t_{k}=0$$ $$y_{k+2}-3y_{k+1}+y_{k}=0$$

$(t_1,y_1)=(3,1)\to(x_1,y_1)=(1,1)$

$(t_2,y_2)=(7,3)\to(x_2,y_2)=(2,3)$

$(t_3,y_3)=(18,8)\to(x_3,y_3)=(5,8)$

$(t_4,y_4)=(47,21)\to(x_4,y_4)=(13,21)$

Since $1, 2, 3, 5, 8, 13, 21$ are in Fibonacci sequence, we predict that $(x_n,y_n)=(F_{2n-1},F_{2n})$

and will prove by Induction.

$(x_n,y_n)=(F_{2n-1},F_{2n})$ is true for $n=1, 2, 3, 4$

Suppose that $(x_k,y_k)=(F_{2k-1},F_{2k})$ is true,

Since $y_{k+1}-3y_k+y_{k-1}=0$, so $y_{k+1}=3y_k-y_{k-1}=3F_{2k}-F_{2k-2}$

$=3F_{2k}-(F_{2k}-F_{2k-1})=2F_{2k}+F_{2k-1}=F_{2k}+F_{2k+1}=F_{2k+2}$

$x_{k+1}=\frac{t_{k+1}-y_{k+1}}{2}=\frac{(3t_k-t_k)-3y_{k-1}-y_{k-1}}{2}$

$=3\left(\frac{t_k-y_k}{2}\right)-\left(\frac{t_k-1-y_{k-1}}{2}\right)=3x_k-x_{k-1}$

Similarly, $x_{k+1}=F_{2k+1}$, so $(x_n,y_n)=(F_{2n-1},F_{2n})$

Answer : $(x_n,y_n)=(F_{2n-1},F_{2n})$

Answer 1

Infinite Solutions:

$$x_k=\frac{1}{2}\left[\frac{5+\sqrt 5}{5} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{5-\sqrt 5}{5} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k\right]$$

$$y_k=\frac{5+3\sqrt 5}{10} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{5-3\sqrt 5}{10} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k$$

But you've done all the hard work yourself,

$$xy+x^2-y^2=1 \ \to$$

$$\left(2x+y\right)^2 - 5y^2 = 4$$

when you complete the square.

So we arrive at a simple pell like equation, one with infinite solutions:

$$\text{let} \ \ \ \ t=2x+y \ \to$$ $$t^2-5y^2=4 \tag{*}\ \ \to$$ $$\left(\frac{3t+5y}{2}\right)^2-5\left(\frac{t+3y}{2}\right)^2=4 \ \to$$ $$\begin{cases} t_{k+1}=\frac{3}{2}t_{k}+\frac{5}{2}y_{k} &&& (1)\\ y_{k+1}=\frac{1}{2}t_{k}+\frac{3}{2}y_{k} &&& (2)\end{cases} \ \ \ \to$$

$\underline{(t,y)\to\left(x=\frac{t-y}{2},y\right)}$

$(3,1) \ \to (1,1)$

$(7,3) \ \to (2,3)$

$(18,8) \ \to (5,8)$

$(47,21) \ \to (13,21)$

$ \vdots$

To solve this system $((1) , (2))$ first we can eliminate $y_{k}$:

$$\frac{3}{2}t_{k+1}-\frac{5}{2}y_{k+1}=t_{k}$$ $$\frac{5}{2}y_{k+1}=\frac{3}{2}t_{k+1}-t_{k}$$

Step the recurrence down, so that you can substitute:

$$\frac{5}{2}y_{k}=\frac{3}{2}t_{k}-t_{k-1}$$

Substitute for $\frac{5}{2}y_{k}$ from $(1)$

$$\left(t_{k+1}-\frac{3}{2}t_{k}\right)=\frac{3}{2}t_{k}-t_{k-1}$$

$$t_{k+1}-3t_{k}+t_{k-1}=0$$ Step back up, and here you have a second order linear recurrence relation in just one variable:

$$t_{k+2}-3t_{k+1}+t_{k}=0$$ $$y_{k+2}-3y_{k+1}+y_{k}=0$$

Indeed, you get the same thing solving for $y$. Although it's the same recurrence relation the first and second values may differ, producing different sequences for $t$ and $y$. You solve one, we'll do $t$ here, and repeat the process to solve the other.

Proceeding with t, anyone who has studied the Fibonacci sequence knows the historical "guess". You "guess" that $t_k=\lambda^k$.

$$\lambda^{k+2}-3\lambda^{k+1}+\lambda^k=0$$ Let's assume $\lambda\neq0$, then divide through by $\lambda^k$: $$\lambda^2-3\lambda+1=0$$ Leads us to $$\lambda_1=\frac{3+\sqrt{5}}{2}$$ $$\lambda_2=\frac{3-\sqrt{5}}{2}$$ We use this to update our guess to something in the span of $\lambda_1^k$ and $\lambda_2^k$:

$$t_k=A\cdot\lambda_1^k+B\cdot\lambda_2^k$$

And we can find out what $A$ and $B$ are by using the first and second solution to $(*)$ which are $(3,1)$ and $(7,3)$; we can let $t_0=3$ and $t_1=7$, thus:

$$\begin{cases} A+B=3 \\ A\cdot \lambda_1 + B \cdot \lambda_2 = 7 \end{cases}$$

This time, with matrix notation, we have that

$$\left(\begin{array}{cc|c} 1&1&3\\ \lambda_1 & \lambda_2 & 7 \end{array}\right) \sim \left(\begin{array}{cc|c} 1 & 0 & \frac{7-3\lambda_2}{\lambda_1 - \lambda_2}\\ 0 & 1 & \frac{3\lambda_1 - 7}{\lambda_1 - \lambda_2} \end{array}\right)$$

Thus $$A=\frac{7-3\lambda_2}{\lambda_1 - \lambda_2}=\frac{3+\sqrt5}{2}$$ $$B=\frac{3\lambda_1 - 7}{\lambda_1 - \lambda_2}=\frac{3-\sqrt5}{2}$$

Repeat the process for $y$, finding the $A$ and $B$ for $y$, and our second order recurrences have been solved:

$$t_k=\frac{3+\sqrt5}{2} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{3-\sqrt5}{2} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k$$

$$y_k=\frac{5+3\sqrt 5}{10} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{5-3\sqrt 5}{10} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k$$

Our original substitutions were

$t=2x+y \implies x_k = \frac{t_k - y_k}{2} $

finally leading us to

$$x_k=\frac{1}{2}\left[\frac{5+\sqrt 5}{5} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{5-\sqrt 5}{5} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k\right]$$

$$y_k=\frac{5+3\sqrt 5}{10} \cdot \left(\frac{3+\sqrt{5}}{2}\right)^k + \frac{5-3\sqrt 5}{10} \cdot \left(\frac{3-\sqrt{5}}{2}\right)^k$$

Answer 2

Hint:

You should keep the cube

$$(x^2+xy-y^2)^3 = 1$$

$$(x^2+xy-y^2)^3 -1^3 = 0$$

You should factorize with the formula $a^3-b^3= (a-b) (a² + ab + b²))$

Edit: Then solve the equations.But as pointed by Alex, $ a^2+a=-1$ has no solution in $ Z $, so only $a=1$ need to be solved.

Answer 3

It's obvious that $(1,1)$ is one of solutions.

Let $y-1=m(x-1)$ be an equation of the line which has other solutions.

Hence, $m\in\mathbb Q$ and after substitution in $x^2+xy-y^2=1$ we obtain: $$x=\frac{m^2-2m+2}{m^2-m-1}$$ and $$y=\frac{2m-1}{m^2-m-1}.$$ I hope it will help.

Answer 4

Cassini's identity for the Fibonacci numbers is $$ F_{n-1}F_{n+1}-F_{n}^{2}=(-1)^{n} $$ Therefore $$ 1 =F_{2n-1}F_{2n+1}-F_{2n}^{2} =F_{2n-1}(F_{2n-1}+F_{2n})-F_{2n}^{2} =F_{2n-1}^2+F_{2n-1}F_{2n}-F_{2n}^{2} $$ and $x=F_{2n-1}$, $y=F_{2n}$ are solutions of $x^2+xy-y^2 = 1$.

Similarly, $$ F_{2n-1}^2-F_{2n-1}F_{2n-2}-F_{2n-2}^{2} = 1 $$ and $x=-F_{2n-1}$, $y=F_{2n-2}$ are solutions of $x^2+xy-y^2 = 1$. But the OP only wants positive solutions.

It remains to prove that these are the only solutions.