I need help solving this Diophantine Equation I have. Its X^2+Y^2+1=3XY
The answers must be integers and the only one I found was x=1, y=1.
I don't quite know what you just said here...
We can narrow down some possible solutions by looking modulo $\displaystyle n $.
Looking modulo $\displaystyle 3 $:
$\displaystyle 3xy\equiv0\mod{3}\implies x^2+y^2+1\equiv0\mod{3}\implies x^2,y^2\equiv1\mod{3}\implies x $ and $\displaystyle y $ can't be multiples of $\displaystyle 3 $.
I'll let you know of any more progress I make.
Break through!
Let's just look at the solutions when $\displaystyle x $ is positive.
Let $\displaystyle \{x_1,x_2,\cdot\cdot\cdot,x_n,\cdot\cdot\cdot\} $ be all ordered $\displaystyle x $ solutions i.e. $\displaystyle x_1=1, x_2=2, x_3=5,... $
I claim the solutions for $\displaystyle y $ paired with $\displaystyle x_n $ is $\displaystyle y=x_{n-1},x_{n+1} $. (Can you see why?)
This is big because now we can find all solutions quite easily!
$\displaystyle x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} $. Let's see if you can find the formula for $\displaystyle x_n $ now.
Precisely! (Well almost: You mean odd-labeled Fibonacci numbers.)
Now I guess all that's left is to show the odd-labeled Fibonacci numbers satisfy $\displaystyle x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} $.
I'm having a lot of trouble showing this though... what's your take on this?
I count from zero!
I haven't tried yet; I've got too many things to do tonight! No good idea comes to mind though. I'll give it a shot tomorrow if I can.
Now I guess all that's left is to show the odd-labeled Fibonacci numbers satisfy $\displaystyle x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} $.
I'm having a lot of trouble showing this though... what's your take on this?
One more thing: you need to show this is the only sequence that generates positive solutions. I've figured out a way, but let's see you try first.
We want to show $\displaystyle F_{2n+3}=\frac{3F_{2n+1}+\sqrt{5F_{2n+1}^2-4}}{2} $.
Now $\displaystyle F_{2n+3} = F_{2n+2}+F_{2n+1} = F_{2n+1}+F_{2n}+F_{2n+1} = 2F_{2n+1}+F_{2n} $.
Therefore it suffices to show $\displaystyle \sqrt{5F_{2n+1}^2-4} = F_{2n+1}+2F_{2n} $. (Can you see why?)
So... $\displaystyle \sqrt{5F_{2n+1}^2-4} = F_{2n+1}+2F_{2n} \iff F_{2n+1}F_{2n}+F_{2n}^2 = F_{2n+1}^2-1 $.
This last result can be shown inductively (I omit the details, but have verified it).
Since we know this is the only sequence of positive integers that satisfies this equation, we have all solutions to $\displaystyle x^2+y^2+1=3xy $ with $\displaystyle x>0 $:
$\displaystyle (x,y) = \{(F_{2n+1},F_{2n-1}),\; (F_{2n+1},F_{2n+3}) \;|\; n\geq0\} $. (When $\displaystyle n=0 $, $\displaystyle F_{2n-1} = F_{-1} = 1 $.)
The solutions with $\displaystyle x<0 $ follow analogously.
I already did use the excel to show the answers for that equation. Since I received this Equation as an assignment, I could really use some help. My teacher pointed me to the direction of assigning Fibonnacci numbers as algebraic expressions. Sorry for the late responses I usually don't have access to a computer regularly.