Let $p(x) ≡ x^3 + ax^2 + bx - 1$, and $q(x) ≡ x^3 + cx^2 + dx + 1$ be polynomials with integer coefficients. Let $α$ be a root of $p(x) = 0$. $p(x)$ is irreducible over the rationals. $α + 1$ is a root of $q(x) = 0$. Find an expression for another root of p(x) = 0 in terms of $α$, but not involving $a$, $b$, $c$, or $d$.
My question is about the following solution:
Let the roots of $p(x)$ be $α$, $β$, $γ$. The polynomial $q(x+1)$ has $α$ as one of its roots. Suppose it is different from $p(x)$. Then by subtracting we get either a quadratic or a linear equation which also has $α$ as a root. It cannot be a linear equation, because then $α$ would be rational and hence $p(x)$ would not be irreducible over the rationals. If it was quadratic, then by multiplying it by a suitable rational factor $rx + s$ and subtracting from $p(x)$ we would get a linear factor $ux + v$ which also had $α$ as a root. If this factor is zero, then we have reduced $p(x)$ over the rationals (as ($rx + s$) times the quadratic). If not, then $α$ is a root of $ux + v$ and hence rational, so that $p(x)$ is still reducible. So we must have that $p(x)$ and $q(x+1)$ are identical.
Rest can be found here: https://prase.cz/kalva/putnam/psoln/psol764.html
Specifically, I'm ok up to the stage of "subtracting from $p(x)$ we would get a linear factor $ux + v$ which also had $α$ as a root..."
Why is a linear factor guaranteed? Perhaps I'm failing to read something into the previous line about the suitable rational factor that would make this happen?
Also, supposing we do get this linear factor, what does it then mean for it to be zero?
We have $$\alpha^3+a\alpha^2+b\alpha-1=0$$ $$(\alpha+1)^3+c(\alpha+1)^2+d(\alpha+1)+1=0$$ from which $$a\alpha^2+b\alpha-1=3\alpha^2+3\alpha+1+c(\alpha+1)^2+d(\alpha+1)+1$$ Because $\alpha$ is an irrational of degree $3$, the coefficients of the powers of $\alpha$ in this last equation should be zero so
$c=a-3,d=a,b=3(a-1)$ then $$p(x)=x^3+ax^2+3(a-1)x-1$$ From $p(\alpha)=0$ we have $$\boxed{a=\frac{3\alpha+1-\alpha^3}{\alpha^2+3\alpha}\Rightarrow b=\frac{-3(\alpha^3+\alpha^2-1)}{\alpha^2+3\alpha}}$$ Now by Vieta we have the system for the other two roots $\alpha_1,\alpha_2$ $$\alpha+\alpha_1+\alpha_2=-a$$ $$\alpha\alpha_1\alpha_2=-(-1)=1$$ Which are solutions of the equation $$X^2+(a+\alpha)X+\frac{1}{\alpha}=0$$ We are done.