Difference between revisions of "1970 Canadian MO Problems/Problem 10"
(→Solution) |
(→Solution) |
||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
Set g(x) = f(x) − 5. Since a, b, c, d are all roots of g(x), we must have | 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) | + | <cmath>g(x) = (x − a) (x − b) (x − c) (x − d) h(x)</cmath> |
for some h(x) ∈ Z[x]. Let k be an integer such that f(k) = 8, giving g(k) = | 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 | f(k) − 5 = 3. Using the factorization above, we find that | ||
Revision as of 16:00, 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
\[g(x) = (x − a) (x − b) (x − c) (x − d) h(x)\] (Error compiling LaTeX. Unknown error_msg)
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
