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?

  • 1
    $\begingroup$ It is false. Consider $P(x)=1$, $Q(x)=0$. $\endgroup$
    – Deif
    Commented May 20 at 22:10
  • $\begingroup$ Perhaps the problem intended non-constant polynomials. What's the exact wording? $\endgroup$
    – Dan
    Commented May 20 at 22:15
  • 2
    $\begingroup$ 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. $\endgroup$
    – zaq
    Commented May 20 at 22:37
  • $\begingroup$ 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. $\endgroup$ Commented May 20 at 22:53

2 Answers 2


$$ P^r-Q^r=\prod_{j=0}^{r-1}\left(P-e^\frac{2\pi ij}{r}Q\right)\ . $$ 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 $\ s=1\ .$

  • $\begingroup$ Accepting this because it's very elegant and probably the intended idea. Thanks! $\endgroup$
    – 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

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

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


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