Difference between revisions of "1987 USAMO Problems/Problem 1"
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− | + | ==Problem== | |
− | + | Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers. | |
− | <cmath>2n^ | + | |
− | + | ==Solution== | |
− | + | Expanding both sides, <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath> | |
− | <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== | ||
+ | |||
+ | {{USAMO box|year=1987|before=First<br>Problem|num-a=2}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 11:53, 24 December 2024
Problem
Find all solutions to , where m and n are non-zero integers.
Solution
Expanding both sides, Note that can be canceled and as , can be factored out. Writing this as a quadratic equation in : . The discriminant equals , 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 .
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.