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Thursday, September 16, 2010

Ternary Polynomial

A polynomial with integer coefficients is called "ternary" if its coefficients are all 1, -1, or zero.

First Question:
Find two polynomials P(x) and Q(x) with integer coefficients where at least one coefficient is larger than 2010, such that P(x)Q(x) is ternary.

Second Question:
Does there exist a ternary multiple of F(x) = x^2 - 3x + 1 ?

Solution

First Question

Note that the polynomial (x-1)(x^2-1)(x^4-1)(x^8-1) ... (x^{2^n}-1) is ternary. Indeed, because the coefficient of x^k can only be formed in exactly one way. For digits in the binary representation of k that is one, we take the x^{2^i} term, and for the other digits, we take the -1 term.

Now, that polynomial can be factored. For example:
(x-1)(x^2-1)(x^4-1)(x^8-1)
=(x-1)^{4} (x+1) (x^3+x^2+x+1) (x^7+...+1)
=(x-1)^{4) (x+1) (x+1)(x^2+1) (x+1)(x^2+1)(x^4+1)
=(x-1)^4 (x+1)^3 (x^2+1)^2 (x^4+1)

So similarly,
(x-1)(x^2-1)(x^4-1)(x^8-1) ... (x^{2^n}-1)
= (x-1)^{n+1} (x+1)^n (x^2+1)^{n-1} ... (x^{2^{n-1}}+1)

If we let P(x) = (x-1)^{n+1} and Q(x) to be the rest, we claim that we can make the largest coefficient in P and Q to be arbitrarily large by setting n large enough. For P, it is clear due to binomial expansion. For Q, the factor (x+1)^n will have a large enough coefficient, while the rest of the factors all consist of positive signs (no negative signs).

So by setting n large enough, we could find P and Q that satisfy the desired conditions.

Second Question

We start by proving a lemma: if s > 2 then s^n > s^{n-1} + ...+ s + 1. This lemma can be proved by induction:
s^{n+1} > 2s^n = s^n + s^n > s^n + ... + s + 1

Now, we note that the larger root of F is greater than 2, so let r be this larger root. ( r = (3 + \sqrt{13})/2 > 2 ).

Suppose P is a ternary polynomial such that P(x) = F(x) G(x) for some integer polynomial G, then P(r) = 0.

But that is impossible if r > 2, due to the lemma above, since the largest term in P cannot be offset by the rest of the terms. A contradiction.

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