Find all integers n≥3 for which the polynomial W(x)=xn−3xn−1+2xn−2+6 can be written as a product of two non-constant polynomials with integer coefficients.
Problem
Source: Czech-Polish-Slovak 2005 Q3
Tags: algebra, polynomial, modular arithmetic, quadratics, algebra unsolved
28.04.2013 04:46
First, the trivial part is to check for linear factors. If my calculations are right, the only possibility is (x+1) could be a factor when n≡1(mod2). Otherwise, assume all factors to be at least quadratic, and let W(x)=f(x)g(x). Work in F2[x]. Then W(x)=xn−1(x−3). Since this is the irreducible in our field, we therefore can write (WLOG) f(x)=xi(x−3)+2h(x),g(x)=xn−1−i+2k(x), where 1≤i≤n−3(because we already checked all linear factors) and deg(h(x))<deg(f(x)) and deg(k(x))<deg(g(x)). Now we see that f(x)g(x)=xn−3xn−1+2h(x)xn−1−i+2k(x)xi(x−3)+4h(x)k(x)=W(x)=xn−3xn−1+2xn−2+6, so after subtracting and dividing by 2, h(x)xn−1−i+k(x)xi(x−3)+2h(x)k(x)=xn−2+3. But since 1≤i≤n−3, we can match constant terms to see that the constant term of 2h(x)k(x) is 3. But this is clearly a contradiction, because 3 is odd. So all odd n work, and all evens don't. Remark: What I essentially did is prove Extended Eisenstein's(Theorem 3).
13.10.2019 18:35
We claim that the answer is all odd integers n≥3. For odd n≥3, x+1 divides W(x) and so odd n≥3 clearly work. Let's now show that even n≥4 do not work. We will work in Z3[x]. Notice that the polynomial W(x) is just xn−2(x2+2) in Z3[x]. Assume, for contradiction, that W(x)=A(x)B(x) for some nonconstant A,B∈Z[x]. WLOG assume that A,B are both monic. Notice that, in Z3[x], we have AB=xn−2(x+1)(x+2). We therefore have one of two cases, neglecting the order of A(x) and B(x). Case 1. For some a+b=n−2, A(x)=(x+1)xa and B(x)=(x+2)xb in Z3[x] If neither of a and b is zero, then the constant term of AB is divisible by 9, clearly absurd. Else, a or b is zero, meaning that a monic linear polynomial divides W(x). This is clearly absurd for n even. Case 2. For some a+b=n−2, A(x)=(x+1)(x+2)xa and B(x)=xb in Z3[x]. Again, one of a,b is zero for similar reasons as before, and so it must be a. Clearly, A(x) must then be of the form x2+3kx+2 or x2+3kx−1. It's easy to see that these don't work if k=0. Also, x2+3x+2 and x2−3x+2 easily don't work. Otherwise, it can be checked by the Quadratic Formula that A(x) has a real root of absolute value greater than 3, say r. If r<0, then rn−3rn−1+2rn−2+6>0 because each term is positive. If r>0, then rn−3rn−1>0 and 2rn−2+6>0. In either case W(r)>0, contradicting the fact that r is a root of W(x). As we've now exhausted all cases, we can conclude that even n≥4 do not work. Thus, the answer is all odd n≥3. ◻
13.11.2019 19:17
Here is my solution for this problem Solution It's easy to see that: n≡1 (mod 2) satisfy the problem With n=4, we have: W(x)=x4−3x3+2x2+6 Since: W(x)>0,∀x∈Z, then: W(x) can't be written as a product of two non-constant polynomials with integer coefficients and odd degrees So: W(x)=(x2+ax+b)(x2+cx+d) Hence: a+c=−3, ac+b+d=2, ad+bc=0, bd=6 These equations have no integer solutions Then: W(x)=x4−3x3+2x2+6 is irreducible With n≥6, suppose: W(x)=Q(x)R(x) with Q(x), R(x) are non-constant polynomials with integer coefficients WLOG, we consider: degQ=k≥n2 Let Q(x)=(x−x1)(x−x2).....(x−xk) with x1,x2,.....,xk∈C So: |Q(0)|=|x1x2.....xk|∈{1;2;3;6} But: W(x1)=W(x2)=.....=W(xk)=0 then: |(Q(0))nQ(1)Q(2)|=6k If |Q(0)|=1, we have: |Q(1)Q(2)|=6k≤36, which is a contradiction since: k≥n2≥3 If |Q(0)|=6, we have: |(Q(0))nQ(1)Q(2)|=6k So: k=n, which is a contradiction since degR>0 If |Q(0)|=2, we have: |Q(1)Q(2)|=3k2n−k, conflict with |Q(1)|, |Q(2)| ∈ Z If |Q(0)|=3, we have: |Q(1)Q(2)|=2k3n−k, conflict with |Q(1)|, |Q(2)| ∈ Z Hence with n≡0 (mod 2), W(x) is irreducible
02.04.2020 21:44
Thank you, khanhnx! Your solution is nice and accessible...