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If we have a reducible quadratic function

\begin{equation*} P(x)=a_1x^2+b_1x+c_1=(rx-x_1)(tx-x_2),~x_1,x_2,r,t\in\mathbb{Z}, \end{equation*}

does there exist another irreducible quadratic function $Q(x)=a_2x^2+b_2x+c_2$ such that for every natural $x$, there exists a natural $y$ such that $Q(x)=P(y)$? For example, is there an irreducible quadratic function $Q(x)$ such that whenever $x\in\mathbb{N}$, $Q(x)$ is equal to

\begin{equation*} P(y)=y^2-1=(y+1)(y-1) \end{equation*}

for some $y\in \mathbb{N}$?

Avi
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    @GeorgeS Please don't edit questions to enforce a particular LaTeX style; for instance, don't edit $$ into \begin{equation*} (e.g. in this question and others). These are trivial edits and bump posts unnecessarily. – apnorton May 12 '15 at 13:59
  • @apnorton In George S's defense, it was previously $ rather than $$, and he may simply think it is more readable this way. – Avi May 12 '15 at 15:22
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    yes, but the prior text was written so inline text made sense. Rewording and re-TeX-ing simply to move $ -> {equation*} is unnecessary at best. – apnorton May 12 '15 at 15:23

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EDIT: This does not answer the question that was asked --- see comments

The answer is no if you restrict to real-valued quadratics (which presumably we can because all complex-valued quadratics are reducible...). You can convince yourself of this by drawing some pictures of quadratics.

Take for example the reducible $P(x)=x^2-1=(x-1)(x+1)$.

Now $P(2)=+3$ and $P(0)=-1$.

Now suppose we have an irreducible $Q(x)$ and $y_1,\,y_2\in\mathbb{R}$ such that $Q(y_1)=P(2)=+3$ and $Q(y_2)=P(x_2)=-1$.

As all quadratics are continuous this implies that $Q$ has a root between $y_1$ and $y_2$ and hence by the Factor Theorem has a factor and is thus reducible.

JP McCarthy
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  • But it doesn't imply that the root is rational; note that we defined "reducible" as factorable into the form $(rx-x_1)(tx-x_2)$ for integers $r,t,x_1,x_2$. – Avi May 12 '15 at 12:04