Author |
Topic: The American Polynomial (Read 526 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
The American Polynomial
« on: Mar 7th, 2003, 11:17am » |
 Quote  Modify
|
What little is known about the American Polynomial can be summarised as follows:
1) It is of a single variable, and has degree at least 2.
2) All of its coefficients are integers.
3) Its constant term is 1492.
4) The sum of the coefficients of its even exponents (including the constant term) is 1776.
5) The sum of the coefficients of its odd exponents is 1621.
Does the American Polynomial have any integer roots?
|
| « Last Edit: Mar 7th, 2003, 11:31am by NickH » |
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
aero_guy
Senior Riddler
Gender:
Posts: 513
|
 |
Re: The American Polynomial
« Reply #1 on: Mar 7th, 2003, 12:48pm » |
Quote Modify
|
A quick analysis of the second order version shows the roots are not integers, therefore it cannot be said that there must be integer roots. The remaining question is to prove that there are no integer roots.
we have:
sum(aix2i)+sum(bix2i-1)+1492=0
sum(ai)=284
sum(bi)=1621
It seems there may be some basic rule of polynomials I don't know here. Maybe it has something to do with 1621 being prime.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: The American Polynomial
« Reply #2 on: Mar 7th, 2003, 3:34pm » |
Quote Modify
|
Does this have anything to do with the Eisenstein Criterion?
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: The American Polynomial
« Reply #4 on: Mar 7th, 2003, 8:56pm » |
Quote Modify
|
How about this:Let P be the polynomial. The three conditions gives
P(0)=1492
(P(1)+P(-1))/2 = 1776
(P(1)-P(-1))/2 = 1621.
Adding and subtracting the latter two gives:
P(1) = 3397 = 43 x 79
P(-1) = 155 = 5 x 71
P(0) = 2 x 2 x 373
Suppose k is an integer for which P(k) = 0. Then P(x) = (x - k)R(x) for some polynomial R. Synthetic division shows that since P(x) has integer coefficients, R must as well, so R(0), R(1), and R(-1) are all integers. Now,
(-k)R(0) = 1492 ==> k divides 1492
(1-k)R(1) = 3397 ==> k-1 divides 3397
(-1-k)R(-1) = 155 ==> k+1 divides 155
An examination of the factorizations above shows that no k satisfies all three conditions.
Hence P can have no integer roots.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
|
Re: The American Polynomial
« Reply #5 on: Mar 8th, 2003, 3:13am » |
Quote Modify
|
Comments on Icarus' solution...
Great solution!
My solution also begins by calculating P(0), P(1), and P(-1), but then takes a slightly different tack. There's a way to reach the conclusion without factorizing the values...
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
 |
Re: The American Polynomial
« Reply #6 on: Mar 24th, 2003, 1:52pm » |
Quote Modify
|
The alternative solution I alluded to above, which also uses the polynomial remainder theorem, goes like this...
3 divides p(3) - p(0), and p(3k) = p(0) (mod 3).
Similarly, p(3k+1) = p(1) (mod 3), p(3k-1) = p(-1) (mod 3).
Since p(-1), p(0), p(1) are all <> 0 (mod 3), the result follows.
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print