Difference between revisions of "1987 USAMO Problems/Problem 1"
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==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. |
| + | |||
| + | Do it | ||
==Solution== | ==Solution== | ||
| − | { | + | Expanding both sides, <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath> |
| + | Note that <math>m^3</math> can be canceled and as <math>n \neq 0</math>, <math>n</math> can be factored out. | ||
| + | Writing this as a quadratic equation in <math>n</math>: <cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</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. | ||
| + | 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 01:27, 11 May 2024
Problem
Find all solutions to 
, where m and n are non-zero integers.
Do it
Solution
Expanding both sides, ![\[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\]](http://latex.artofproblemsolving.com/7/6/7/767ed379169cd1698227a5c3701945e4848daaa1.png)
Note that 
can be canceled and as 
, 
can be factored out.
Writing this as a quadratic equation in : ![\[2n^2+(m^2-3m)n+(3m^2+m)=0\]](http://latex.artofproblemsolving.com/9/7/5/975eb800f6268d629e943923140e0a35c9af5afe.png)
.
The discriminant 
equals ![\[(m^2-3m)^2-8(3m^2+m)\]](http://latex.artofproblemsolving.com/b/7/4/b74901a1eaa73af8d79e69b33369fa4f5fa9467b.png)
![\[=m^4-6m^3-15m^2-8m\]](http://latex.artofproblemsolving.com/2/8/9/289612920e3842a5bffee2f9ee58a5849d1f6d47.png)
, which we want to be a perfect square.
Miraculously, this factors as 
. This is square iff (if and only if) 
is square or 
. It can be checked that the only nonzero 
that work are 
. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs 
as ![\[\{(-1,-1),(8,-10),(9,-6),(9,-21)\}\]](http://latex.artofproblemsolving.com/0/2/a/02ab2e52a1c531ede30d7228036dbdbf7e1a0f36.png)
.
See Also
| 1987 USAMO (Problems • Resources) | ||
| 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. 
