Difference between revisions of "1976 USAMO Problems/Problem 5"
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so <math>(x-1)</math> is a factor of <math>P(x)</math>, as desired. <math>\blacksquare</math> | so <math>(x-1)</math> is a factor of <math>P(x)</math>, as desired. <math>\blacksquare</math> | ||
+ | === Solution 3 === | ||
+ | Let <math>z, z^2, z^3</math> be three of the 5th roots of unity not equal to one that satisfy <math>1 + z + z^2 + z^3 + z^4 = 0</math> as a result. Plugging them into the equation gives the linear system of equations in <math>(A(1), B(1), C(1))</math>: | ||
+ | <cmath>A(1) + zB(1) + z^2C(1) = 0</cmath> | ||
+ | <cmath>A(1) + z^2B(1) + z^4C(1) = 0</cmath> | ||
+ | <cmath>A(1) + z^3B(1) + z^6C(1) = 0</cmath> | ||
+ | Direct observation gives <math>(A(1), B(1), C(1)) = (0,0,0)</math> as a solution, and there are no others because the system of equations is linear. Hence, <math>A(1) = 0</math>, and so <math>(x-1)</math> is a factor of <math>A(x)</math>, as desired. <math>\blacksquare</math> | ||
− | + | ''Note: We can generalize this approach to prove the claim in Solution 1 in an equally fast, concise, and readily understandable fashion.'' | |
== See Also == | == See Also == |
Latest revision as of 17:14, 16 August 2015
Problem
If , , , and are all polynomials such that prove that is a factor of .
Solutions
Solution 1
In general we will show that if is an integer less than and and are polynomials satisfying then , for all integers . For the problem, we may set , , and then note that since , is a factor of .
Indeed, let be the th roots of unity other than 1. Then for all integers , for all integers . This means that the th degree polynomial has distinct roots. Therefore all its coefficients must be zero, so for all integers , as desired.
Solution 2
Let be three distinct primitive fifth roots of unity. Setting , we have These equations imply that or But by symmetry, Since , it follows that . Then, as noted above, so is a factor of , as desired.
Solution 3
Let be three of the 5th roots of unity not equal to one that satisfy as a result. Plugging them into the equation gives the linear system of equations in :
Direct observation gives as a solution, and there are no others because the system of equations is linear. Hence, , and so is a factor of , as desired.
Note: We can generalize this approach to prove the claim in Solution 1 in an equally fast, concise, and readily understandable fashion.
See Also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
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.