Difference between revisions of "1987 USAMO Problems/Problem 1"
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==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 <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>. | |
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==See Also== | ==See Also== | ||
Revision as of 20:57, 10 April 2018
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 is square. 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 |
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