# Polynomials - rational root - divisibility

This is a simple (almost obvious) question but I somehow cannot find an elementary solution yet.
$$f(x) = a_0x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n$$
is a polynomial with integer coefficients.

$$\frac{p}{q}$$ is a rational root of $$f(x)$$,
where $$p,q$$ are whole numbers and $$(p,q) = 1$$

Prove that $$(p-mq)$$ | $$f(m)$$ for every whole number $$m$$

Assume the solution has to be explained to someone in 7th or 8th grade
i.e. one who doesn't know much about any complicated theory of polynomials.

E.g. I know that $$f(x)$$ can be written as $$a_0\left(x-\frac{p}{q}\right)R(x)$$
where $$R(x)$$ is a polynomial but can I claim $$R(x)$$ has integer
coefficients. If I can, then the desired statement follows easily, it seems.
But why can I claim this?

Note: you can assume I (or the student to whom I must explain this)
know that $$p/a_n$$ and $$q/a_0$$, this was proved already.

At some point I thought of using induction (on $$m$$) but that
seems somewhat complicated and I don't know if it will work at all.

• I am somewhat confused by the fact that the $x^k$ term doesn't have $a_k$ as coefficient. – Arthur Feb 14 at 9:50
• @Arthur OK, well... that's how it's in my book. Doesn't matter. You can let $b_k = a_{n-k}$ – peter.petrov Feb 14 at 9:51
• Note that the book offers no solution or even hint for this statement :) – peter.petrov Feb 14 at 10:06
• This is a less used, imo, notation for polynomials. I think most authors would prefer nowadays to write down polynomials where the index of each coefficient mathces the power of $\;x\;$ . This is specially easier to grasp for high school students. – DonAntonio Feb 14 at 10:08
• @DonAntonio I don't know, I've been taught about polynomials with this "reverse" notation and this book also uses it. It doesn't really matter. – peter.petrov Feb 14 at 10:10

Suppose $$R = x^{n-1} + \dots + b_{1}x + b_{0}$$, then $$f = a_0 \left(x - \frac{p}{q} \right) R = a_0 x^n + \left(a_0 b_{n-2} - a_0 \frac{p}{q} \right) x^{n-1} + \left(a_0 b_{n-3} - \frac{p}{q} \right) x^{n-2} + \dots + \left(-a_0 b_0 \frac{p}{q} \right).$$ So $$a_0 b_{n-k} - a_0 \frac{p}{q} \in \mathbb{Z}$$. Thus, the $$a_0 b_{n-k} \in \frac{1}{q} \mathbb{Z}$$. On the other hand, $$\operatorname{gcd}(q,p-mq) = 1$$, so since $$a_0 R \in \frac{1}{q} \mathbb{Z}[x]$$, your desired result follows.

• I will need to digest this a bit. Thanks, will reread this a few more times, and see if I can understand it. – peter.petrov Feb 14 at 10:41
• What is $b_k$ here? Is it what I suggested $b_k = a_{n-k}$ ? – peter.petrov Feb 14 at 11:18
• @peter.petrov Those are just the coefficients of the polynomial $R$. Sorry, I didn't see that you used the notation $b$ in the comments already. – Qi Zhu Feb 14 at 11:29

We will use induction on $$n$$ (the degree of $$f$$)
for the statement we are trying to prove which is

$$(p-mq)$$ | $$f(m)$$ for every whole number $$m$$

The base case $$n=1$$ is very easy. So let's assume we have this already.

We know that $$q\ |\ a_0 \tag{1}$$

So let's denote $$a_0 = q \cdot c_0 \tag{2}$$

Note that $$c_0$$ is a whole number (as are all $$a_i$$).

OK... then we get:

$$f(x) = a_0x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n =$$

$$= c_0 q x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n =$$

$$= (xq - p) c_0 x^{n-1} + (a_1 + pc_0)x^{n-1} + a_2 x^{n-2} + ... + a_{n-1} x + a_n \tag{3}$$

Now we know that $$p/q$$ is a root of $$f(x)$$ and obviously it is also a root of the first term $$(xq - p) c_0 x^{n-1}$$ from the sum above.

So $$p/q$$ must also be a root of

$$h(x) = (a_1 + pc_0)x^{n-1} + a_2 x^{n-2} + ... + a_{n-1} x + a_n$$

But this polynomial $$h(x)$$ is of degree $$(n-1)$$ so we can apply the induction hypothesis. From it we get that $$(p-mq)\ |\ h(m) \tag{4}$$ for every whole number $$m$$.

But then from (3) we can see that $$f(m) = (mq - p) \cdot m^{n-1} + h(m) \tag{5}$$
for every whole number $$m$$.

And then from (4) and (5), we obtain that
$$(p-mq)$$ divides also $$f(m)$$ for every whole number $$m$$.

Is this proof good enough? Or does this proof have some subtle flaw?

If it's good enough, then the statement is indeed easy because everything was straightforward in the induction that we did, and we used just the fact that $$q\ |\ a_0$$.