$(a)$ Prove that for every positive integer $n$, the number of ordered pairs $(x, y)$ of integers satisfying $x^2-xy+y^2 = n$ is divisible by $3.$ $(b)$ Find all ordered pairs of integers satisfying $x^2-xy+y^2=727.$
Problem
Source: Turkey TST 2000 P1
Tags: algebra, polynomial, Vieta, number theory unsolved, number theory
mathmdmb
12.03.2011 12:48
Potla wrote: $(a)$ Prove that for every positive integer $n$, the number of ordered pairs $(x, y)$ of integers satisfying $x^2-xy+y^2 = n$ is divisible by $3.$ $(b)$ Find all ordered pairs of integers satisfying $x^2-xy+y^2=727.$ (a).Note that if $(x,y)$ is a solution,then so is $(x,-y),(-x,y),(-x,-y),(x,x-y),(y,y-x)$.So in fact the number of solutions is divisible by $6$
From Vietta,if we fix $x$ first let $y,y_1$ be the two solutions.
And then $y_1=x-y$
Use Vietta and the discriminant is a perfect square
Rijul saini
12.03.2011 15:31
mathmdmb wrote: Potla wrote: $(a)$ Prove that for every positive integer $n$, the number of ordered pairs $(x, y)$ of integers satisfying $x^2-xy+y^2 = n$ is divisible by $3.$ $(b)$ Find all ordered pairs of integers satisfying $x^2-xy+y^2=727.$ (a).Note that if $(x,y)$ is a solution,then so is $(x,-y),(-x,y),(-x,-y),(x,x-y),(y,y-x)$.So in fact the number of solutions is divisible by $6$ The first two don't actually work.
Anyways, I think we can say that $(x,y)$ (where we can take WLOG that $x$ is positive and greater than $y$) is a solution implies $(-x,-y),(y,x),(-y,-x),(x,x-y),(x-y,x)$ are other solutions and these cover all solutions.
The case $x=y$ can be handled easily.
mathmdmb
13.03.2011 06:14
Rijul saini wrote: mathmdmb wrote: Potla wrote: $(a)$ Prove that for every positive integer $n$, the number of ordered pairs $(x, y)$ of integers satisfying $x^2-xy+y^2 = n$ is divisible by $3.$ $(b)$ Find all ordered pairs of integers satisfying $x^2-xy+y^2=727.$ (a).Note that if $(x,y)$ is a solution,then so is $(x,-y),(-x,y),(-x,-y),(x,x-y),(y,y-x)$.So in fact the number of solutions is divisible by $6$ The first two don't actually work.
Anyways, I think we can say that $(x,y)$ (where we can take WLOG that $x$ is positive and greater than $y$) is a solution implies $(-x,-y),(y,x),(-y,-x),(x,x-y),(x-y,x)$ are other solutions and these cover all solutions.
The case $x=y$ can be handled easily.
Great careless mistake
Cogent
13.03.2011 14:43
Potla wrote: $(a)$ Prove that for every positive integer $n$, the number of ordered pairs $(x, y)$ of integers satisfying $x^2-xy+y^2 = n$ is divisible by $3.$ $(b)$ Find all ordered pairs of integers satisfying $x^2-xy+y^2=727.$ (a) If $(x,y)$ where $x\neq y$ is a solution, then the other solutions are $(y,x),(-x,-y),(-y,-x),(x,x-y),(-x,y-x),$ $(y,y-x),(-y,x-y),(x-y,x),(y-x,-x),(x-y,-y),(y-x,y)$. If $x=y$, then there are six solutions only. (b) (13,31) is a solution. Other solutions are (31,13), (-13,-31), (-31,-13), (13,-18), (-13,18), (31,18), (-31,-18), (-18,13), (18,-13), (-18,-31), (18,31).
$x^2-xy+y^2=727.$
$\Rightarrow x^2+y^2+(x-y)^2=2*727$
mathmdmb
14.03.2011 07:05
Complete solution for (b) Let wlog $x\le y$,since obviously $x\neq y,x<y$ Let it be fixed on $y$ and it is enough to find solutions in positive integers,then we can find all solutions from (a) Its discriminant $2908-3y^2\ge 0\implies y^2\le 961\implies y\le 31$ On the other hand $x^2-xy+y^2<y^2\implies 727<y^2\implies y\ge 27$ That is $27\le y\le 31$ Now just check with $y=27,28,29,30,31$ which is rather easy to see that only $y=31$ works.Then $x^2-31x+234=0\implies x=18,13$