Difference between revisions of "1987 USAMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | + | Simply both sides completely <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath> | |
− | + | Canceling out like terms gives us <cmath>mn+m^2n^2+n^3=-3m^2n+3mn^2-n^3</cmath> | |
+ | Moving <math>2n^3</math> to the RHS and factoring out the <math>3n</math> gives us <cmath>n(m+m^2n-n^2)=3n(-m^2+mn-n^2)</cmath> | ||
+ | Only nonzero solutions are needed, so <math>n</math> can be divided off. <cmath>m+m^2n-n^2=3(-m^2+mn-n^2)</cmath> | ||
+ | Move all terms with factors of <math>n</math> to the RHS and simplifying <cmath>m=-3m^2+3mn-2n^2-m^2n</cmath> | ||
+ | We should remove the cubic term. ....... | ||
==See Also== | ==See Also== | ||
Revision as of 13:27, 1 April 2017
Problem
Find all solutions to , where m and n are non-zero integers.
Solution
Simply both sides completely
Canceling out like terms gives us
Moving
to the RHS and factoring out the
gives us
Only nonzero solutions are needed, so
can be divided off.
Move all terms with factors of
to the RHS and simplifying
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.