# An elementary olympiad polynomial problem.

If $$r, s \in \mathbb{N}; r > s$$ are such that $$\mathrm{gcd}(r, s) = 1$$ and there exist different nonconstant real polynomials $$P, Q$$ that satisfy

$$(P(x))^r - (Q(x))^r = (P(x))^s - (Q(x))^s; \forall x \in \mathbb{R},$$

show that $$r = 2, s = 1$$.

I have tried some of the typical approaches; e.g. I tried analyzing the roots of $$P$$ and $$Q$$, but it seems too ugly. I have also tried solving the case $$s = 1$$ first, but even that is very hard and I haven't been able to do it. Any ideas?

• It is false. Consider $P(x)=1$, $Q(x)=0$.
– Deif
Commented May 20 at 22:10
• Perhaps the problem intended non-constant polynomials. What's the exact wording?
– Dan
Commented May 20 at 22:15
• Sorry, I was translating and I did forget the condition that $r, s$ are coprime. I'm not sure why, but it's not required in the original problem that the polynomials are nonconstant. I will add it in as this is clearly not what was intended.
– zaq
Commented May 20 at 22:37
• Try to divide $P^r-Q^r$ by $P^s-Q^s$ and see where you end up. The coprime conditions allows you to write $1= a\cdot r + b\cdot s$ with some integers $a,b\in \mathbb{Z}.$ This could help, too. Commented May 20 at 22:53

$$P^r-Q^r=\prod_{j=0}^{r-1}\left(P-e^\frac{2\pi ij}{r}Q\right)\ .$$ Therefore, since $$\ P\ne Q\ ,$$ $$P^r-Q^r=P^s-Q^s\Rightarrow\prod_{j=1}^{r-1}\left(P-e^\frac{2\pi ij}{r}Q\right)=\prod_{j=1}^{s-1}\left(P-e^\frac{2\pi ij}{s}Q\right)\ .$$ Since $$\ P,Q\$$ are real non-constant polynomials, $$\ P- e^{i\theta}Q\$$ cannot be a constant polynomial for any real $$\ \theta\in(0,2\pi)\$$ except $$\ \theta=\pi\ .$$ Therefore it's only possible for $$\ \prod_\limits{j=1}^{r-1}\left(P-e^\frac{2\pi ij}{r}Q\right)\$$ and $$\ \prod_\limits{j=1}^{s-1}\left(P-e^\frac{2\pi ij}{s}Q\right)\$$ to have the same degree if $$\ r=s+1\$$ and is even (whence $$\ s\$$ must be odd), and $$\ P+Q=c\ ,$$ a constant polynomial. In that case $$(c-Q)^{s+1}-Q^{s+1}=(c-Q)^s-Q^s\ ,\tag{1}\label{e1}$$ or $$c^{s+1}\equiv c^s\pmod{Q}\ ,$$ from which it follows that $$\ c=0\$$ or $$\ c=1\$$ because $$\ Q\$$ is a non-constant polynomial. The first alternative is impossible, because it gives $$P^r-Q^r=(-Q)^r-Q^r=0\ ,$$ and $$P^s-Q^s=(-Q)^s-Q^s=-2Q^s\ne0\ .$$ Therefore equation \eqref{e1} becomes $$(1-Q)^{s+1}-Q^{s+1}=(1-Q)^s-Q^s\ ,$$ or $$(1-Q)^s(-Q)=Q^s(Q-1)\ .$$ Since $$\ Q\ne0\$$ and $$\ Q\ne1\ ,$$ it follows that $$(1-Q)^{s-1}=Q^{s-1}\ ,$$ which is only possible if $$\ s=1\ .$$

• Accepting this because it's very elegant and probably the intended idea. Thanks!
– zaq
Commented May 21 at 12:22

The polynomial relation is

$$(P(x))^r - (Q(x))^r = (P(x))^s - (Q(x))^s \tag{1}\label{eq1A}$$

Since $$r \gt s$$, plus both $$P(x)$$ and $$Q(x)$$ are not constant, the degree of the highest order term among $$(P(x))^r$$ and $$(Q(x))^r$$ is larger than that among $$(P(x))^s$$ and $$(Q(x))^s$$. Thus, for the two sides in \eqref{eq1A} to be equal, the highest order terms among $$(P(x))^r$$ and $$(Q(x))^r$$ must be equal in degree, call it $$n$$, and cancel each other. Thus, with

$$P(x) = \sum_{i=0}^{n}p_{i}x^{i}, \;\; Q(x) = \sum_{i=0}^{n}q_{i}x^{i} \tag{2}\label{eq2A}$$

we have

$$p_n^{r} - q_n^{r} = 0 \;\;\to\;\; p_n^{r} = q_n^{r} \tag{3}\label{eq3A}$$

If $$r$$ is odd, then $$p_n = q_n$$, while if $$r$$ is even, we have $$p_n = \pm q_n$$. Next, factoring out $$P(x) - Q(x)$$ from both sides of \eqref{eq1A}, and using that this is not the zero polynomial so we can divide by it, we get

$$P(x)^{r-1} + P(x)^{r-2}Q(x) + \ldots + Q(x)^{r-1} = P(x)^{s-1} + P(x)^{s-2}Q(x) + \ldots + Q(x)^{s-1} \tag{4}\label{eq4A}$$

The highest order terms on the LHS has an exponent of $$(r-1)n$$ while the RHS ones are $$(s-1)n$$, which is smaller, so the sum of the LHS $$(r-1)n$$ exponent terms must be $$0$$. However, if $$p_n = q_n$$, then this exponent is $$n(q_n)^{r-1}$$ instead, which is not $$0$$. Thus, $$r$$ must be even, so $$s$$ is odd as $$\gcd(r, s) = 1$$, and also

$$p_n = -q_n \tag{5}\label{eq5A}$$

Next, define the polynomial

$$T(x) = P(x) + Q(x) \;\;\to\;\; P(x) = -Q(x) + T(x) \tag{6}\label{eq6A}$$

Note \eqref{eq5A} means the degree of $$T(x)$$, call it $$m$$, is less than $$n$$, but it's not the $$0$$ polynomial since, if it were, then the LHS of \eqref{eq1A} would be $$0$$ while the RHS would be $$-2(Q(x))^s$$ instead. Substituting \eqref{eq6A} into \eqref{eq1A}, expanding using the Binomial theorem, and using that $$r$$ is even while $$s$$ is odd, we get

$$r(-Q(x))^{r-1}T(x) + \frac{r(r-1)}{2}(-Q(x))^{r-2}(T(x))^2 + \frac{r(r-1)(r-2)}{6}(-Q(x))^{r-3}(T(x))^3 + \ldots + (T(x))^r \\ = -2(Q(x))^{s} + s(-Q(x))^{s-1}T(x) + \frac{s(s-1)}{2}(-Q(x))^{s-2}(T(x))^2 + \ldots + (T(x))^s \tag{7}\label{eq7A}$$

Due to $$m \lt n$$, the highest order terms on each side come from the first terms shown. Thus, their degrees must match, giving that

$$(r - 1)n + m = sn \;\to\; (r - s - 1)n + m = 0 \tag{8}\label{eq8A}$$

However, since $$r \gt s$$ and $$n \gt 0$$, then $$(r - s - 1)n \ge 0$$, plus also $$m \ge 0$$, so we must have

$$r = s + 1, \;\; m = 0 \tag{9}\label{eq9A}$$

Thus, $$T(x)$$ is a constant, call it $$T_1$$. Next, the coefficients of the terms must also match, which gives that

$$-rT_1 = -2 \;\;\to\;\; \frac{rT_1}{2} = 1 \tag{10}\label{eq10A}$$

The exponents of the second and third terms on each side in \eqref{eq7A} also match, and since they are the highest ones, then their coefficients must also be the same. With the second terms, we just get $$r = s + 1$$ again, but using \eqref{eq9A} and \eqref{eq10A} with the third term coefficients results in

\begin{aligned} \left(\frac{r(r-1)(r-2}{6}\right)T_1^3 & = \left(\frac{s(s-1)}{2}\right)T_1^2 \\ \left(\frac{rT_1}{2}\right)\frac{(r-1)(r-2)}{3} & = \frac{s(s-1)}{2} \\ 2((s+1)-1)((s+1)-2) & = 3s(s-1) \\ 2s(s-1) & = 3s(s-1) \\ 0 & = s(s-1) \end{aligned}\tag{11}\label{eq11A}

Since $$s$$ is a positive integer, this means that $$s = 1$$, so from \eqref{eq9A}, we get that $$r = 2$$.