Find a pair of polynomials $a(x)$ and $b(x)$ so that $a(x)+b(x)=x^6 -1$ and the $\gcd(a(x),b(x))=x+1$.

Question

Could there be infinite number of answers?

Answer 1

We have $$\gcd(a(x),x^6-1)=\gcd(a(x),a(x)+b(x))=\gcd(a(x),b(x))=x+1$$ Hence $(x+1)|a(x)$ and similarly $(x+1)|b(x)$. We write $a(x)=(x+1)\alpha$ and $b(x)=(x+1)\beta$. We now factor $x^6-1=(x+1)(x^5-x^4+x^3-x^2+x-1)$.

Now we seek $\alpha+\beta=x^5-x^4+x^3-x^2+x-1$ and $\gcd(\alpha,\beta)=1$. For example, take $\alpha=x^5, \beta=-x^4+x^3-x^2+x-1$, which gives $$a(x)=(x+1)x^5,~~ b(x)=(x+1)(-x^4+x^3-x^2+x-1)$$

By choosing a monomial $\gamma$ carefully, we can get infinitely many solutions by taking $\alpha=x^5+\gamma$, $\beta=-\gamma-x^4+x^3-x^2+x-1$.