Difference between revisions of "1987 USAMO Problems/Problem 1"

m (remove "do it")
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers.
+
Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers.  
  
 
==Solution==
 
==Solution==
Line 8: Line 8:
 
The discriminant <math>b^2-4ac</math> equals <cmath>(m^2-3m)^2-8(3m^2+m)</cmath>
 
The discriminant <math>b^2-4ac</math> equals <cmath>(m^2-3m)^2-8(3m^2+m)</cmath>
 
<cmath>=m^4-6m^3-15m^2-8m</cmath>, which we want to be a perfect square.
 
<cmath>=m^4-6m^3-15m^2-8m</cmath>, which we want to be a perfect square.
Miraculously, this factors as <math>m(m-8)(m+1)^2</math>. This is square iff <math>m^2-8m</math> is square. It can be checked that the only nonzero <math>m</math> that work are <math>-1, 8, 9</math>. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs <math>(m, n)</math> as <cmath>\{(-1,-1),(8,-10),(9,-6),(9,-21)\}</cmath>.
+
Miraculously, this factors as <math>m(m-8)(m+1)^2</math>. This is square iff (if and only if) <math>m^2-8m</math> is square or <math>m+1=0</math>. It can be checked that the only nonzero <math>m</math> that work are <math>-1, 8, 9</math>. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs <math>(m, n)</math> as <cmath>\{(-1,-1),(8,-10),(9,-6),(9,-21)\}</cmath>.
  
 
==See Also==
 
==See Also==

Latest revision as of 11:53, 24 December 2024

Problem

Find all solutions to $(m^2+n)(m + n^2)= (m - n)^3$, where m and n are non-zero integers.

Solution

Expanding both sides, \[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\] Note that $m^3$ can be canceled and as $n \neq 0$, $n$ can be factored out. Writing this as a quadratic equation in $n$: \[2n^2+(m^2-3m)n+(3m^2+m)=0\]. The discriminant $b^2-4ac$ equals \[(m^2-3m)^2-8(3m^2+m)\] \[=m^4-6m^3-15m^2-8m\], which we want to be a perfect square. Miraculously, this factors as $m(m-8)(m+1)^2$. This is square iff (if and only if) $m^2-8m$ is square or $m+1=0$. It can be checked that the only nonzero $m$ that work are $-1, 8, 9$. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs $(m, n)$ as \[\{(-1,-1),(8,-10),(9,-6),(9,-21)\}\].

See Also

1987 USAMO (ProblemsResources)
Preceded by
First
Problem
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png