Prove that : $\gcd ( a.P(X) + b.Q(X), cP(X) + dQ(X) ) = \gcd ( P(X), Q(X) )$

Question

Let $P(X)$ and $Q(X)$ be two polynomials in $R[X].$ Let $a, b, c, d$ be four real numbers such that $ad – bc ≠ 0$. Prove that :
$$\gcd(aP(X) + bQ(X), cP(X) + dQ(X))= \gcd (P(X), Q(X)).$$

Can someone help me to resolve this ? I have no idea how to do it.

Answer 1

Let $$r(x)= \gcd ( a.P(X) + b.Q(X), cP(X) + dQ(X) )$$ and $$s(x) = \gcd ( P(X), Q(X) ).$$

Since $P=sP'$ and $Q= sQ'$ we have $$aP+bQ = s(aP'+bQ')$$ and $$cP+dQ = s(cP'+dQ')$$ which means that $s\mid aP+bQ$ and $s\mid cP+dQ$ so $\boxed{s\mid r}$.

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Vice versa, we prove $r\mid s$ (and so $s=r$): $$r\mid c(aP+bQ)-a(cP+dQ) = (cb-ad)Q $$

so $r\mid Q$ and the same way we get $r\mid P$ so $\boxed{r\mid s}$

Answer 2

Let $\tilde P(X)=aP(X) + bQ(X)$ and $\tilde Q(X)=cP(X) + dQ(X)$. Then $$ \begin{pmatrix} \tilde P(X) \\ \tilde Q(X) \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} P(X) \\ Q(X) \end{pmatrix} $$ and so $$ \begin{pmatrix} P(X) \\ Q(X) \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} \begin{pmatrix} \tilde P(X) \\ \tilde Q(X) \end{pmatrix} $$ because $ad – bc \ne 0$. Therefore $\gcd(\tilde P(X,\tilde Q(X))$ divides $\gcd (P(X), Q(X))$.