Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1982 IMO Problems/Problem 4: Difference between revisions

Cosinator (talk | contribs)
91 edits
page/ problem/ solution
 
Temperal (talk | contribs)
3,154 edits
format
 
Line 1: Line 1:
==Problem==
Prove that if <math>n</math> is a positive integer such that the equation <math>x^3-3xy^2+y^3=n</math> has a solution in integers <math>x,y</math>, then it has at least three such solutions. Show that the equation has no solutions in integers for <math>n=2891</math>.
Prove that if <math>n</math> is a positive integer such that the equation <math>x^3-3xy^2+y^3=n</math> has a solution in integers <math>x,y</math>, then it has at least three such solutions. Show that the equation has no solutions in integers for <math>n=2891</math>.


==Solution==
Suppose the equation <math>x^3-3xy^2+y^3=n</math> has solution in integers <math>(x,y)</math> with <math>y=x+k</math>.  Then, completing the cube yields <math>(y-x)^3-3x^2y+2x^3</math>.  Using the substitution <math>y=x+k</math> yields <math>k^3-3kx^2-x^3=n</math>.  Notice that equality directly implies that <math>(k,-x)</math> is also a solution to the original equation.  Applying the transformation again yields that <math>(-x-k,-k)</math> is also a solution.  We show that these three solutions are indeed distinct: If <math>(x,y)=(k,-x)</math> then <math>x=k,x+k=-x</math> which only has solution <math>x=k=0</math> which implies that <math>n</math> is not a positive integer, a contradiction.  Similarly, since the transformation from <math>(k,-x)</math> to <math>(-x-k,-k)</math> and <math>(-x-k,-k)</math> to <math>(x,y)</math> is the same as the transformation from <math>(x,y)</math> to <math>(k,-x)</math>, we have that the three solutions are pairwise distinct.
Suppose the equation <math>x^3-3xy^2+y^3=n</math> has solution in integers <math>(x,y)</math> with <math>y=x+k</math>.  Then, completing the cube yields <math>(y-x)^3-3x^2y+2x^3</math>.  Using the substitution <math>y=x+k</math> yields <math>k^3-3kx^2-x^3=n</math>.  Notice that equality directly implies that <math>(k,-x)</math> is also a solution to the original equation.  Applying the transformation again yields that <math>(-x-k,-k)</math> is also a solution.  We show that these three solutions are indeed distinct: If <math>(x,y)=(k,-x)</math> then <math>x=k,x+k=-x</math> which only has solution <math>x=k=0</math> which implies that <math>n</math> is not a positive integer, a contradiction.  Similarly, since the transformation from <math>(k,-x)</math> to <math>(-x-k,-k)</math> and <math>(-x-k,-k)</math> to <math>(x,y)</math> is the same as the transformation from <math>(x,y)</math> to <math>(k,-x)</math>, we have that the three solutions are pairwise distinct.


For the case <math>n=2891</math>, notice that <math>7|n</math>.  Considering all solutions modulo <math>7</math> of the equation <math>x^3-3xy^2+y^3\equiv0\pmod{7}</math> yields only <math>x\equiv y\equiv 0\pmod{7}</math>.  But, this implies that <math>7^3</math> divides <math>2891</math> which is clearly not true.
For the case <math>n=2891</math>, notice that <math>7|n</math>.  Considering all solutions modulo <math>7</math> of the equation <math>x^3-3xy^2+y^3\equiv0\pmod{7}</math> yields only <math>x\equiv y\equiv 0\pmod{7}</math>.  But, this implies that <math>7^3</math> divides <math>2891</math> which is clearly not true.
{{IMO box|year=1982|num-b=3|num-a=5}}
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 21:52, 7 December 2008

Problem

Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.

Solution

Suppose the equation $x^3-3xy^2+y^3=n$ has solution in integers $(x,y)$ with $y=x+k$. Then, completing the cube yields $(y-x)^3-3x^2y+2x^3$. Using the substitution $y=x+k$ yields $k^3-3kx^2-x^3=n$. Notice that equality directly implies that $(k,-x)$ is also a solution to the original equation. Applying the transformation again yields that $(-x-k,-k)$ is also a solution. We show that these three solutions are indeed distinct: If $(x,y)=(k,-x)$ then $x=k,x+k=-x$ which only has solution $x=k=0$ which implies that $n$ is not a positive integer, a contradiction. Similarly, since the transformation from $(k,-x)$ to $(-x-k,-k)$ and $(-x-k,-k)$ to $(x,y)$ is the same as the transformation from $(x,y)$ to $(k,-x)$, we have that the three solutions are pairwise distinct.

For the case $n=2891$, notice that $7|n$. Considering all solutions modulo $7$ of the equation $x^3-3xy^2+y^3\equiv0\pmod{7}$ yields only $x\equiv y\equiv 0\pmod{7}$. But, this implies that $7^3$ divides $2891$ which is clearly not true.

1982 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1982_IMO_Problems/Problem_4&oldid=28553"