The question is to prove that the polynomial $P(x)$ is of even degree if there exists a polynomial $Q(x)$ such that $P(x^2+1) = P(x)Q(x) $. This was an interview question. It is easy to show that $P(x)$ can't be linear and that the degree of $P(x)$ and $Q(x)$ must be equal, also the coefficient of the highest degree term in $Q(x)$ must be $1$. I am not able to proceed further. Any help is appreciated.
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3Do the polynomials need to have real coefficients? – Sambo Jun 12 '23 at 11:55
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1Assuming you have a real coefficient, consider $x^2 + 1 = p$, where $p$ is the root of the $P(x)$. – openspace Jun 12 '23 at 12:01
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Yes the polynomial is supposed to have real coefficients – saket kumar Jun 12 '23 at 12:23
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1Though considering it was a maths interview, it should have been mentioned – saket kumar Jun 12 '23 at 12:25
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1@AspiringMat , I should have mentioned the college lol. It's called ISI(Indian Statistical Institute) – saket kumar Jul 13 '23 at 18:51
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As Dietrich Burde points out, there are counterexamples when you consider polynomials with complex coefficients. On the other hand, if you meant $P(x)$ is a polynomial with real coefficients, then it is indeed true that it must have even degree. This will follow from proving that $P$ has no real roots. Use the fact that if $\alpha$ is a root of $P$, then so is $f(\alpha) = \alpha^2+1$. Now, for any real root $\alpha$, you can keep applying $f$ to get new roots. Prove that this process will yield infinitely many distinct roots for $P$, which would be a contradiction.
Haran
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This is false, as the following example with a linear polynomial $P(x)$ shows, namely $$ P(x)=ax+1,\; Q(x)=x+b,\; \text{ with } a=\frac{\sqrt{-3}-1}{2}, b=\frac{\sqrt{-3}+1}{2}. $$ Then $a+1=b$ so that $$ P(x^2+1)=ax^2+b=P(x)Q(x). $$
Sambo
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Dietrich Burde
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The statement only holds if we insist the coefficients of $P(x)$ are real (counterexample below).
Let's assume $P(x)$ is a real polynomial with degree $n$, where $n$ is odd. We know that all such $P$ have at least one real root, so let $w$ be a real root of $P$. But then $$P(w^2+1)=P(w)Q(w)=0$$ so $w^2+1$ is also a real root of $P$; and it can't be the same real root as $$w^2+1=w$$ doesn't have any real solutions.
But now we can just create the sequence $$w_1=w,\quad w_{k+1}=w_k^2+1$$
which clearly has infinitely many distinct real terms (since $w_{k+1}>w_k$ for all $k$), all of which are roots of $P$; but $P$ has at most $n$ real roots; contradiction.
So $P$ can't be a real polynomial with odd degree. (This is technically enough to answer the question as it's written; but it's also easy enough to find a polynomial $P$ with even degree that works).
If polynomials with complex coefficients are allowed, then $$P(x)=x+c,\quad Q(x)=x-c$$ where $c=\frac{-1+i\sqrt{3}}{2}$ satisfies the condition.
Chris Lewis
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