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Topic: Take One Polynomial, Get Integer Coefficients? (Read 359 times) |
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K Sengupta
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Take One Polynomial, Get Integer Coefficients?
« on: Jul 24th, 2007, 7:45am » |
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Consider the polynomial L(m) = ms + 5*ms-1 + 3, where s is a positive whole number >=2.
Analytically determine whether L(m) can be expressed as the product of two non-constant polynomials having integer coefficients.
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| « Last Edit: Jul 24th, 2007, 9:51am by K Sengupta » |
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Eigenray
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Re: Take One Polynomial, Get Integer Coeffic
« Reply #1 on: Jul 24th, 2007, 3:24pm » |
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The first part uses complex analysis:
| hidden: | Let f(x)=xs-1(x+5). Since |L(x)-f(x)|=3 < 4  |x+5| = |f(x)| for |x|=1, it follows from Rouche's theorem that L has the same number of roots inside the unit disc, counting multiplicity, that f does, namely (s-1). Since the product of the roots of L is  3, the remaining root must be outside the unit disc. |
(Is there an elementary proof?) Now,
| hidden: | | if L factors, then since the constant term 3 is prime, one factor must have a constant term of 1. But this means that there is some subset of the roots of L which multiply to 1. Since one root has norm > 1, and the rest < 1, and the product of the norms is 3, this is impossible (unless we take the empty set, which corresponds to a trivial factorization). |
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Barukh
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Re: Take One Polynomial, Get Integer Coefficients
« Reply #2 on: Jul 25th, 2007, 12:55am » |
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Wow! I am sure a more elementary solution exists.
I tried to apply two knows (to me) irreducibility criteria (Eisenstein and enough-prime-numbers evaluation). Both got me to nowhere... 
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K Sengupta
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Re: Take One Polynomial, Get Integer Coefficients
« Reply #3 on: Jul 27th, 2007, 8:40am » |
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I submit herewith a proposed solution to the given problem.
Still hiding it, because I am not entirely sure about the veracity of the undernoted methodology.
| hidden: | Let L(m)
= (mp + cp-1mp-1+.....+c1m +/- 3) (mq + dq-1mp-1+.....+d1m +/- 1)
We will show that all c's are divisible by 3 and use that fact to establich a contradiction.
At the outset, both p and q must be >=2. If possible, let p=1. Then, it follows that +/-3 is a root. Accordingly, if s is even, we would have:
0 = 3s(+/-3 +5) + 3 which is false since +/- 3 + 5 = 8 or -2.
Similarly, if s is odd, we would have:
0 = 3s-1(+/-3 +5) +3, which is false since +/-3 + 5 = 8 or 2.
If q=1, then +/-1 is a root and we would reach a contradiction in a similar manner.
Hence, p<= s-2 and accordingly, the coefficients of m, m2, ......., mp are all zero.
So, we must have c1 +/- 3d1 = 0, and accordingly c1 is divisible by 3.
We will prove the remaining part by way of induction.
Let us assume that c1, ...., cv are all divisible by 3. We consider the coefficient of mv+1.
If q-1 > = v+1, then :
cv+1 corresponds to a linear combination of c1, ....., cv-1, cv +/- 3dv+1
If p-1< v+1, then cv+1 corresponds to a linear combination of some or all of c1, c2, ......, cv.
In both the above cases, cv+1 is divisible by 3.
Accordingly, considering the coefficients of m, m2, ....., mp-1, we observe that all the c's are multiples of 3.
Now, let us consider the coefficient of ms, which is also zero. It is a sum of the terms which are multiples of 3 added to +/-1, and so, the said coefficient cannot be zero.
This leads to a contradiction.
Consequently, the factorization is not possible. |
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| « Last Edit: Jul 27th, 2007, 8:46am by K Sengupta » |
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