determine all values
Determine all values that $\dfrac {x^2+y^2+1} {xy+1}$ can take in $\mathbb N$
One one thing is neutral in the universe, that is $0$.
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leonardo shawon
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Re: determine all values
for n, n+1 where n,n+1 € N the answer is 2.
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: determine all values
Apparently, that's one solution. Can you prove that there is no other solution?
That's why we encourage to come with a complete solution or at least some idea that can generate solutions. I mean I'd appreciate more if you even came with some random divisibility stuffs.
That's why we encourage to come with a complete solution or at least some idea that can generate solutions. I mean I'd appreciate more if you even came with some random divisibility stuffs.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
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Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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the arrivals
- Posts:41
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Re: determine all values
perhaps very renowned problem of imo,very passionate problem of imo as far as i m concerned.
a student of quinsland however solve it with induction making introduction to parabolic family....
a student of quinsland however solve it with induction making introduction to parabolic family....
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: determine all values
Did you mean that it is an IMO problem?It is not an IMO problemthe arrivals wrote:perhaps very renowned problem of imo,very passionate problem of imo as far as i m concerned.
a student of quinsland however solve it with induction making introduction to parabolic family....
I think you are saying this $xy|x^2+y^2+1,$prove that $\dfrac {x^2+y^2+1} {xy}=3$
One one thing is neutral in the universe, that is $0$.
Re: determine all values
^Yup...He probably thought about the famous Vieta Jumping problem.
BTW to learn more about vieta Jumping: http://www.georgmohr.dk/tr/tr09taltvieta.pdf
Solution:
WLOG $x>y$ and $x-y=d$
\[ \frac{x^2+y^2+1}{xy+1}=\frac{(x-y)^2-1+2(xy+1)}{xy+1}=\frac{d^2-1}{xy+1}+2\]
So, $xy+1|d^2-1$.
If $d^2-1 \not = 0$, then $d^2-1 \geq xy+1 \iff d^2 \geq x(d+x)+2=dx+x^2+2$.
But $x\geq d$. So, $d=1$ and $(x,y)=(n+1,n)$.
BTW to learn more about vieta Jumping: http://www.georgmohr.dk/tr/tr09taltvieta.pdf
Solution:
WLOG $x>y$ and $x-y=d$
\[ \frac{x^2+y^2+1}{xy+1}=\frac{(x-y)^2-1+2(xy+1)}{xy+1}=\frac{d^2-1}{xy+1}+2\]
So, $xy+1|d^2-1$.
If $d^2-1 \not = 0$, then $d^2-1 \geq xy+1 \iff d^2 \geq x(d+x)+2=dx+x^2+2$.
But $x\geq d$. So, $d=1$ and $(x,y)=(n+1,n)$.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: determine all values
But please note that I asked the value of the fraction which is $2$
And simply use Vietta Jumping.
And simply use Vietta Jumping.
One one thing is neutral in the universe, that is $0$.
Re: determine all values
It will be $dy+y^2$Moon wrote: If $d^2-1 \not = 0$, then $d^2-1 \geq xy+1 \iff d^2 \geq \boxed{x(d+x)+2=dx+x^2+2}$.
One one thing is neutral in the universe, that is $0$.