An impossible integral polynomial:
Prove that there are no integers a, b, c, d such that the
polynomial P(x) = ax^3+bx^2+cx+d has P(19)=1, P(62)=2.
** HINT 1 **
Consider the "casting out 9s" divisibility test.
** HINT 2 **
Consider casting out 9s in the form: 9 divides P(10)-P(1).
** SPOILER **
The result holds for any integral polynomial P, i.e. any P in Z[x].
The high-school Polynomial Remainder Theorem [PRT] implies
x-y | P(x)-P(y) in Z[x,y], where "a | b" means "a divides b"
i.e. P(x)-P(y) = (x-y) Q(x,y) for some integral polynomial Q
x=62,y=10 => 43=62-19 | P(62)-P(19)=1, i.e. 43|1, contradiction.
More familiar: x=10,y=1 yields the "casting out nines" test
9 | P(10)-P(1) i.e. mod 9: N = P(10) = P(1) = S = digit sum
since decimal notation is a polynomial function of the radix 10:
N = P(10) = a + b 10 + c 10^2 + d 10^3 + ...
S = P(1) = a + b + c + d + ... = sum of digits.
[PRT], by linearity, reduces to the cyclotomic case
x - y | x^n - y^n for all n in N
which has an immediate inductive proof considering
n+1 n+1 n n
x - y n x - y
---------- = x + y -------
x - y x - y
For further applications and generalizations etc.
see my numerous related posts [1].
-Bill Dubuque
[1] http://www.dejanews.com/dnquery.xp?QRY=dubuque%20polynomial%20remainder&groups=sci.math&ST=PS
--
From - Wed Sep 2 14:38:46 1998
From: Bill Dubuque