1987 USAMO Problems/Problem 1
Problem
Find all solutions to 
, where m and n are non-zero integers.
Solution
Simply both sides completely ![\[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\]](http://latex.artofproblemsolving.com/7/6/7/767ed379169cd1698227a5c3701945e4848daaa1.png)
Canceling out like terms gives us ![\[mn+m^2n^2+n^3=-3m^2n+3mn^2-n^3\]](http://latex.artofproblemsolving.com/5/b/7/5b76a44cfef92e1272b93609ab3f6d42dd7edac6.png)
Moving 
to the RHS and factoring out the 
gives us ![\[n(m+m^2n-n^2)=3n(-m^2+mn-n^2)\]](http://latex.artofproblemsolving.com/8/3/8/838732e587aa261b7ada60eaf4fa7be5382aeae5.png)
Only nonzero solutions are needed, so 
can be divided off. ![\[m+m^2n-n^2=3(-m^2+mn-n^2)\]](http://latex.artofproblemsolving.com/4/0/f/40f064bfd61ebcf6687f28bdc0cd4f19d37a70c8.png)
Move all terms with factors of to the RHS and simplifying ![\[m-3m^2=3mn-2n^2-m^2n\]](http://latex.artofproblemsolving.com/6/1/e/61e65b7131f5dfc6dca15f9e9d210f80b1d9cf2e.png)
We should remove the cubic term. .......
See Also
| 1987 USAMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
