Difference between revisions of "1970 Canadian MO Problems/Problem 10"
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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 | ||
Latest revision as of 16:04, 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
