answer

sol

What's the motivation for the above solution?

@above $P(x)=\frac{x^6+1}{x^2+1}*(x+1)=x^5 + x^4 - x^3 - x^2 + x + 1 $ $Q(x)=\frac{x^3+1}{x+1}*(x^2+1)=x^4 - x^3 + 2 x^2 - x + 1 $ Then $P(x)Q(x)=(x^3+1)(x^6+1)=x^9+x^6+x^3+1$ Other example $P(x)=x^2-x+1$ $Q(x)=\frac{x^9+1}{x^2-x+1}=x^7+x^6-x^4-x^3+x+1$

Is there a polynomial of degree <9 with positive coefficients that can be factored as the product of two mixed polynomials? If so, what's the minimum degree example?

$(100x^4+100x^3-x^2+100x+100)^2=10000 x^8 + 20000 x^7 + 9800 x^6 + 19800 x^5 + 40001 x^4 + 19800 x^3 + 9800 x^2 + 20000 x + 10000$ $(100x^3+100x^2-x+100)(100x^3-x^2+100x+100)=10000 x^6 + 9900 x^5 + 9800 x^4 + 30001 x^3 + 9800 x^2 + 9900 x + 10000$

If $f=gh$ for polynomials $f,g,h$ with $f$ having non-negative coefficients and $g,h$ mixed, then the minimum degree of $f$ is $6$. Example: $4x^6+2x^5+13x^3+2x+4 = (2x^3+2x^2-x+2)(2x^3-x^2+2x+2)$. For the proof, we can divide through by leading coefficients to make all polynomials monic, and similarly we can divide by any factors of $x$ to make all polynomials have non-zero constant term. We use the notation $[x^k]f$ to denote the coefficient of $x^k$ in $f$. Claim: $\deg f = n+1, \deg g = n, \deg h = 1$ is impossible. Proof: For $h$ mixed, we need $h = x-b$ for some $b>0$. Let $g = x^n + a_{n-1}x^{n-1} + \dots + a_1x+a_0$. $[x^0]f=-ba_0$, so $a_0<0$. $[x^k]f=a_{k-1}-ba_k$ for $k=1,\dots,n-1$, so $a_{k-1} \geq ba_k$. $[x^n]f=a_{n-1}-b$ so $a_{n-1} \geq b$. Chaining these together gives $0>a_0\geq ba_1 \geq \dots \geq b^ka_k \geq \dots \geq b^{n-1}a_{n-1} \geq b^n$, impossible as $b>0$. Claim: $\deg f = 4, \deg g = 2, \deg h = 2$ is impossible. Proof: Let $g = x^2 + ax + b, h = x^2 + cx + d$. Then $[x^0]f = bd$, so either $b,d>0$ or $b,d<0$. If $b,d>0$, then for $g,h$ to be mixed we must have $a,c<0$. But then $[x^3]f = a+c < 0$, impossible. If $b,d<0$, then $[x^2]f = b+d+ac \geq 0$ so $ac>0$. If $a,c<0$, then we get a contradiction as above. If $a,c>0$, then $[x^1]f = bc+ad <0$, impossible. Claim: $\deg f = 5, \deg g = 2, \deg h = 3$ is impossible. Proof: Let $g = x^2 + ax + b, h = x^3 + cx^2 + dx+e$. Then $[x^0]f = be$, so either $b,e>0$ or $b,e<0$. If $b,e>0$, then by $g$ mixed, $a<0$. As $[x^4]f=a+c$, we must have $c>0$ so by $h$ mixed, $d<0$. But then $[x^1]f=ae+bd<0$, impossible. If $b,e<0$, we look at the following 6 cases: 1. If $a=0$, then $[x^1]f = bd$ so $d \leq 0$. But then $[x^3]f = b+d <0$, impossible. 2. If $d=0$, then $[x^1]f = ae$ so $a \leq 0$. Also, $[x^2]f = e+bc$ so $c<0$. But then $[x^4]f = a+c <0$, impossible. 3. If $a,d<0$, then as $[x^4]f = a+c$, we have $c>0$. But then $[x^3]f = b+d+ac <0$, impossible. 4. If $a,d>0$, then $[x^1]f = ae+bd <0$, impossible. 5. If $a>0, d<0$, then $[x^3]f = b+d+ac$ gives $c>0$. But then $[x^2]f = e+ad+bc <0$, impossible. 6. If $a<0, d>0$, then $[x^4]f = a+c$ gives $c>0$. But then $[x^2]f = e+ad+bc <0$, impossible. Together, these three claims are enough to rule out $\deg f \leq 5$, so we must have $\deg f \geq 6$.