# $P(x)\in \mathbb{R}[x]$, If $P(x)=n$ has at least one rational zero for $\forall n \in \mathbb{N}$, $P(x)=ax+b$

Question:

Let $$P(x)\in \mathbb{R}[x]$$. If $$P(x)=n$$ has at least one rational zero for $$\forall n \in \mathbb{N}$$, show that $$P(x)=ax+b$$ where $$a$$ and $$b$$ are rational.

I totally do not know how I should solve this.

When I set $$g(x)=P(x)-n$$, the derivation is $$g'(x)=P'(x)$$. ... That's what I found. In addition, how can I use the condition that $$P(x)=n$$ has a rational zero? Could you please give me some key points to the problem I'm suffering from? Thanks.

• $g(\xi)=0$ does not imply $g'(\xi)=0$. – Gae. S. Feb 9 at 14:12
• Looks interesting. Where is that problem from ? Do you know for sure that it has a simple solution ? It would also help to know a little about your level of knowledge. – Ewan Delanoy Feb 9 at 14:15
• @Gae.S. I corrected that part. Thanks for your advice! – ToBY Feb 9 at 14:16
• @EwanDelanoy Some Iranian competition? – Aqua Feb 9 at 14:22

Let $$P(x)=\alpha_nx^n+\ldots +\alpha_0$$ with $$\alpha_i\in\Bbb R$$ and $$\alpha_n\ne 0$$. For $$k\in\Bbb N$$, let $$q_k$$ be a rational root of $$P(x)-k$$. Then the $$\deg P+1$$ points $$(q_k,k)$$, $$k=1,2,\ldots, \deg P+1$$ determine $$P$$ uniquely and we can find the coefficients of $$P$$ from them by solving a system of linear equations with all coefficients rational. It follows that all $$\alpha_i$$ are rational. Then for suitable $$M\in\Bbb N$$, all $$a_i:=M\alpha_i$$ are integer. By the rational root theorem, any rational root $$q_k$$ of $$MP(x)-kM$$, i.e., of $$a_nx^2n+a_{n-1}x^{n-1}+\ldots +(a_0-kM)$$ (at least when $$kM\ne a_0$$) is of the form $$q_k=\frac {u_k}{v_k}$$ where $$u_k\mid a_0-kM$$ and $$v_k\mid a_n$$. In particular, $$r_k:=a_nq_k\in \Bbb Z$$ for all $$k$$. (If $$kM=a_0$$, we can divide out a power of $$x$$ and at least still obtain $$q_k\in\frac1{a_n}\Bbb Z$$, so still $$r_k\in\Bbb Z$$). So with $$Q(x):=MP(a_n x)\in\Bbb Z[x]$$ we are given that $$Q(x)-kM$$ has an integer root $$r_k$$ for all $$k\in\Bbb N$$.

By the Mean Value Theorem, there exists $$\xi_k$$ between $$r_k$$ and $$r_{k+1}$$ with $$\left|Q'(\xi_k)\right|=\left|\frac{Q(r_{k+1})-Q(r_k)}{r_{k+1}-r_k}\right|=\frac{M}{|r_{k+1}-r_k|}\le M.$$ Similarly, there exists $$\eta_k$$ between $$r_k$$ and $$r_{k+2}$$ with $$\left|Q'(\eta_k)\right|=\left|\frac{Q(r_{k+2})-Q(r_k)}{r_{k+2}-r_k}\right|=\frac{2M}{|r_{k+2}-r_k|}\le 2M.$$

Assume $$Q'$$ is not constant. Then there exists $$L$$ such that $$|Q(x)|>2M$$ for all $$x$$ with $$|x|>L$$. Therefore $$|\xi_k|\le L$$ and $$|\eta_k|\le L$$ for all $$k$$. Since the $$r_k$$ are all distinct and only finitely many integers are between $$-L$$ and $$L$$, there exists $$k$$ such that $$|r_k|, |r_{k+1}|, |r_{k+2}|$$ are all $$>L$$. As among $$r_k,r_{k+1},r_{k+2}$$, two numbers must have the same sign, it follows that one of $$|\xi_k|, |\xi_{k+1}\|, |\eta_k|$$ is $$>L$$, contradiction. We conclude that $$Q'$$ is constant.

Then $$Q$$, as well as $$P$$ is at most linear. It can clearly not be constant, hence $$P(x)=ax+b$$ with $$a,b\in\Bbb Q$$.

WLOG, suppose that the leading coefficient of $$P$$ is positive. By the Lagrange interpolation formula, $$P$$ must have rational coefficients.

Let $$Q(x) = m P(x)$$, where $$m$$ is a positive integer chosen so that $$Q$$ has integer coefficients. Let the leading coefficient of $$Q$$ be $$c$$, and let its constant term be $$d$$. From the given hypothesis, for all sufficiently large primes $$p$$, we can find some positive rational number $$r$$ such that $$Q(r) = p + d$$.

By the rational root theorem, all rational roots of the polynomial $$Q(x) - (p-d)$$ must be of the form $$\dfrac{\pm p}{s}$$, where $$s$$ is a divisor of $$c$$. Hence, $$r \geq \dfrac{p}{c}$$. On the other hand, if $$\deg Q > 1$$, then

$$0 = Q(r) - (p-d) \geq Q(\frac{p}{c}) - (p-d) \geq O(p^2),$$

which is clearly impossible. Hence, the degree of $$P$$ must be less than or equal to 1.

• How can you suppose that the leading coefficient is positive? Thank you – Giuseppe Negro Feb 9 at 18:52
• @GiuseppeNegro, if it's not positive, you can let $Q(x)=-mP(x)$. – LHF Feb 9 at 18:54
• And will $Q$ satisfy that $Q(x)=n$ has a rational zero for all $n\in\mathbb N$? Is this obvious? Sorry if it is a silly question – Giuseppe Negro Feb 9 at 18:56
• @GiuseppeNegro, It's very good you ask :) $Q(x)$ does not satisfy $Q(x) = n$, that's $P(x)$. $Q(x)$ is the 'common-denominator' version of $P(x)$ with a positive leading coefficient. – LHF Feb 9 at 19:09