1970 Canadian MO Problems/Problem 10: Difference between revisions
Created page with "== Problem == Given the polynomial <math>f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n</math> with integer coefficients <math>a_1,a_2,\ldots,a_n</math>, and given also ..." |
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== Solution == | == Solution == | ||
Set g(x) = f(x) − 5. Since a, b, c, d are all roots of g(x), we must have | |||
g(x) = (x − a) (x − b) (x − c) (x − d) h(x) | |||
for some h(x) ∈ Z[x]. Let k be an integer such that f(k) = 8, giving g(k) = | |||
f(k) − 5 = 3. Using the factorization above, we find that | |||
3 = (k − a) (k − b) (k − c) (k − d) h(x). | |||
By the Fundamental Theorem of Arithmetic, we can only express 3 as the product | |||
of at most three distinct integers (−3, 1, −1). Since k − a, k − b, k − c, k − d | |||
are all distinct integers, we have too many terms in the product, leading to a | |||
contradiction. | |||
via Justin Stevens | |||
Revision as of 15:59, 23 July 2019
Problem
Given the polynomial 
with integer coefficients 
, and given also that there exist four distinct integers 
and 
such that 
, show that there is no integer 
such that 
.
Solution
Set g(x) = f(x) − 5. Since a, b, c, d are all roots of g(x), we must have g(x) = (x − a) (x − b) (x − c) (x − d) h(x) for some h(x) ∈ Z[x]. Let k be an integer such that f(k) = 8, giving g(k) = f(k) − 5 = 3. Using the factorization above, we find that 3 = (k − a) (k − b) (k − c) (k − d) h(x). By the Fundamental Theorem of Arithmetic, we can only express 3 as the product of at most three distinct integers (−3, 1, −1). Since k − a, k − b, k − c, k − d are all distinct integers, we have too many terms in the product, leading to a contradiction.
via Justin Stevens
