That's too easy I think.

Use first derivative

Denote $S_{f}$ as the solution set of $f(x)=0$. From the statement we know that $S_P \cup S_{P+1} = S_Q \cup S_{Q+1}$. Since $10 \cdot \text{deg} P = 21 \cdot \text{deg} Q$, we have $\text{deg} P = 21x$ and $\text{deg} Q = 10x$. Since $|S_{f}| \le \text{deg} f$ for polynomials $f$, we get that $|S_P| + |S_{P+1}| = |S_Q| + |S_{Q+1}| \le 20x$. Now, here's a key lemma that solves the problem. Lemma. For nonconstant polynomial $P$, $|S_P| + |S_{P+1}| \ge \text{deg} P +1$. Proof of Lemma. Denote $P(x) = \prod_{i=1}^n (x-r_i)^{c_i}$ and $P(x)+1 = \prod_{i=1}^m (x-r_i')^{c_i'}$. Clearly $r_i$ and $r'_i$ are pairwise distinct. Now look at $P'(x)$. This must have $r_i$ as a root with multiplicity $c_i-1$, and $r_i'$ as a root with multiplicity $c_i'-1$. This implies that $$ \text{deg} P-1 = \text{deg} P' \ge \sum_{i=1}^n (c_i-1) + \sum_{i=1}^m (c_i'-1) = 2 \cdot \text{deg} P - |S_P| - |S_{P+1}| $$Rearranging gives the desired lemma. Now we have $20x \ge |S_P| + |S_{P+1}| \ge \text{deg} P +1 =21x+1$, an obvious contradiction.

The answer is no; here is the official solution. Assume $(P,Q)$ work, so $\deg P = 21n$ and $\deg Q = 10n$ for some $n$. Then we have: \begin{align*} 1 &= \gcd(P,10P+9) = \gcd(P+1, 10P+9) \\ \implies 1 &= \gcd\left( P^{10} + P^9, 10P+9 \right) = \gcd\left( Q^{21} + Q^{20}, 10P+9 \right) \\ \implies 1 &= \gcd(Q, 10P+9). \end{align*} Now, we take a derivative: \[ P' \cdot P^8 \cdot \left( 10P + 9 \right) = Q' \cdot Q^{19} \cdot \left( 21Q + 20 \right) \neq 0. \]Thus $10P+9$ should divide $Q' \cdot (21Q+20)$ but \[ \deg(10P+9) = 21n > 20n-1 = \deg(Q' \cdot (21Q+20)). \]

For brevity, I will denote $P(x) = P, Q(x) =Q$. Note that the set of roots of $P$ is disjoint with that of $P+1$. Similarly for $Q$. Note that $P' = (P+1)'$, so the number of roots of $P'$, counting multiplicity must not be less than the sum of multiplicities of the roots of $P, P+1$ decreased by 1, so the number of roots of $P$ and $P+1$ is at least $\text{deg} P + 1$. But the number of roots of $P^{9} (P+1)$ doesn't exceed $\frac{20}{21} \deg P $, which is a contradiction.

Using derivative in polynomial has once presented https://artofproblemsolving.com/community/c6h1352166p7389123

If $a = X^{10} + X^9$ and $c = X^{21} + X^{20}$, the equation rewrites as $a \cdot p = c \cdot q$. It is clear that the polynomial $a$ is indecomposable. Then according to Corollary 2.18 in http://dept.math.lsa.umich.edu/~zieve/papers/peter.pdf (which is proven using monodromy groups) there exists a linear polynomial $\ell$ such that $\ell \cdot a = T_{10}$ or $\ell \cdot a = X^{10}$, where $T_{10}$ is the $10$th Chebyshev polynomial. The first case is impossible since $T_{10}$ contains a nontrivial coefficient of $x^8$ but not of $x^9$, and the second case is impossible since $a(-1) = a(0)$ but $(-1)^{10} \neq 0^{10}$.

I think I have an elementary solution for the following more general problem(although I'm not 100% sure) Let $k>l>2$ be two coprime integers. Prove that there are no non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients such that $$P^{k+1}+P^k=Q^{l+1}+Q^l.$$ Even if this general problem is not true it would be interesting to find for which pairs $(k,l)$ the problem is true.

huricane wrote: Even if this general problem is not true it would be interesting to find for which pairs $(k,l)$ the problem is true. The official solution, for instance, proves this for $k \ge 1, l \ge 2k+1$.

Hamel wrote: That's too easy I think. Eh, no.

WizardMath wrote: huricane wrote: Even if this general problem is not true it would be interesting to find for which pairs $(k,l)$ the problem is true. The official solution, for instance, proves this for $k \ge 1, l \ge 2k+1$. Yeah but their method wouldn't work for the general problem, since they use degree bounds.

v_Enhance wrote: Hamel wrote: That's too easy I think. Eh, no. Yeah right. Were it to give with distinct roots it would be easier.

İs it famous method to use derivative in polynomials? So anyone who have seen USA TST 2017 problem could easily do this. Does anyone know any handout/book about usage of derivative in polynomials or some problems about that topic other than USA TST and this one?

The idea isn't really that uncommon,i mean any time you want some further information about roots or just kill the constant it seems like the right thing to do.For much older reference check this (problem 9) Apparently Putnam 1957 B7 wrote: Problem let $S(P)$ be the set of the roots of $P$ not counting multiplicity.Is it possible that $S(P)\equiv S(Q)$ and $S(P+1)\equiv S(Q+1)$ ? Or even a more well know result Mason's theorem ,where the proof essentially goes along the lines of writing $\tfrac{a}{c}+\tfrac{b}{c}=1$ and differentiating to kill the one on the LHS.

Lamp909 wrote: Use first derivative Isn't calculus "loathed" in olympiads.? I mean can you clarify when can one use or not use calculus to solve a problem?

Edit: Wrong Sol

The key statement $|S_P|+|S_{P+1}|\ge \deg P+1$ is same as IMC 2000 day 2, prob 3. See http://imc-math.org.uk/imc2000/prob_sol2.pdf The author was Marianna Csörnyei.

Here's my solution: I invoke notation from post #5. Then we have $|S_P| + |S_{P+1}| = |S_Q| + |S_{Q+1}| \le 20x$, where $\text{deg} (P) = 21x$ and $\text{deg} (Q) = 10x$ as shown in rkm0959's solution. Now, we look at the Lemma given below (it is basically a generalization of the Lemma in post #5), after which we can finish in a similar fashion as in rkm0959's solution. LEMMA: Let $F \in \mathbb{R}[x]$ be a non-constant real valued polynomial of degree $n$ $(n \geq 1)$. Consider the $m+1$ distinct real numbers $r_1,r_2, \dots ,r_{m+1}$ where $m \in \mathbb{N}$. Then the total number of complex solutions to the equation $F(x)=r_i$ for all $i \in \{1,2, \dots ,m+1\}$ is at least $mn+1$. (The previous Lemma follows on taking $m=1$ and $r_1=0,r_2=-1$) Proof of Lemma We always use $i$ to denote the integers from $1$ to $m+1$. First consider a general polynomial $A \in \mathbb{R}[x]$. Suppose that $\gcd(A,A')$ has degree $s$, where $A'$ is the first derivative of $A$. Then there exist $s$ roots of $A$ (not necessarily distinct) which are roots of $A'$ also, which means that the remaining roots of $A$ are distinct and correspond to the elements of $S_A$ (Just use the fact that roots of $A$ with multiplicity $a$ have multiplicity $a-1$ in $A'$). So we have $|S_A|=\text{deg}(A)-s$. Return to our Lemma. Letting $X$ denote the total number of solutions to the given equations, and using the fact that $(F-r_i)'=F'$, we have $$X=\sum_{i=1}^{m+1} n-\text{deg} (\gcd(F-r_i,F'))=(m+1)n-\sum_{i=1}^{m+1} \text{deg}(\gcd(F-r_i,F'))$$Then we wish to show that $X \geq mn+1$, which is equivalent to showing that $$(m+1)n-\sum_{i=1}^{m+1} \text{deg}(\gcd(F-r_i,F')) \geq mn+1 \Leftrightarrow \sum_{i=1}^{m+1} \text{deg}(\gcd(F-r_i,F')) \leq n-1 \text{ } \forall m \in \mathbb{N}$$Consider the polynomial $G(x)=(F(x)-r_1)(F(x)-r_2) \dots (F(x)-r_{m+1})$. As all these terms in product are pairwise co-prime, so we get that $$\sum_{i=1}^{m+1} \text{deg}(\gcd(F-r_i,F'))=\text{deg}(\gcd(G,F')) \leq \text{deg}(F') \leq \deg(F)-1=n-1 \quad \blacksquare$$ REMARK: The Lemma is tbh just a mixture of some well-known facts. For example, the fact that the number of distinct roots of a polynomial $A$ is $\text{deg}(A)-\text{deg}(\gcd(A,A'))$ is well-known (maybe not exactly in that form). Similarly, the second half of the proof of the Lemma (introducing $G$) is well motivated if one catches hold of the fact that degrees get added on multiplying pairwise co-prime polynomials.

Haha derivative go brrr. We get \begin{align*} P(x)^9 \left(P(x) + 1 \right) &= Q(x)^{20} \left(Q(x) + 1 \right) \\ P'(x) \cdot P(x)^8 \left(10P(x) + 9 \right) &= Q'(x) \cdot Q(x)^{19} \left(21Q(x) + 20 \right) \end{align*}to work with. By analyzing the leading term of both sides in the first equation, and because $\gcd(10, 21) = 1$, we find that the degree of $P$ is $21c$ for some $c$ and the degree of $Q$ is $10c$. Observe that $Q(x)$ divides $P(x)(P(x) + 1)$. Additionally, notice that \begin{align*} \gcd(10P(x) + 9, P(x) + 1) = \gcd(10P(x) + 9, P(x)) = 1, \end{align*}hence we know that $10P(x) + 9$ divides $Q'(x) \cdot (21Q(x) + 20)$. But this is cap; the degree of the left is $21c$ and the degree of the right is $10c - 1 + 10c = 20c - 1$, smaller. Thus we are done.

v_Enhance wrote: The answer is no; here is the official solution. Assume $(P,Q)$ work, so $\deg P = 21n$ and $\deg Q = 10n$ for some $n$. Then we have: \begin{align*} 1 &= \gcd(P,10P+9) = \gcd(P+1, 10P+9) \\ \implies 1 &= \gcd\left( P^{10} + P^9, 10P+9 \right) = \gcd\left( Q^{21} + Q^{20}, 10P+9 \right) \\ \implies 1 &= \gcd(Q, 10P+9). \end{align*} Now, we take a derivative: \[ P' \cdot P^8 \cdot \left( 10P + 9 \right) = Q' \cdot Q^{19} \cdot \left( 21Q + 20 \right) \neq 0. \]Thus $10P+9$ should divide $Q' \cdot (21Q+20)$ but \[ \deg(10P+9) = 21n > 20n-1 = \deg(Q' \cdot (21Q+20)). \] What does gcd (P,Q) mean? I never see the definition of greatest common divisor for polynomial.Can you explain,plz?

A bit overcomplicated but the idea is the same. The answer is no. Suppose $r$ is an arbitrary root of either side of the equation (we will refer to this as a "root of the equation"), and let $(A,B)$ denote $\gcd(A,B)$. We consider four cases—note that $r$ must fall in exactly one of these. If $r$ is a root of $(P,Q)$, then it is a root of $P(x)^{10}+P(x)^9$ with multiplicity divisible by $9$, and a root of $Q(x)^{21}+Q(x)^{20}$ with multiplicity divisible by $20$. Thus $r$ is a root of the equation, it must have multiplicity divisible by $180$. Thus we can say that there are $A$ distinct roots of the equation which contribute a total multiplicity of $180a$. If $r$ is a root of $(P+1,Q)$, then it is a root of the equation with multiplicity divisible by $20$, so we can say there are $B$ distinct roots with total multiplicity $20b$. If $r$ is a root of $(P,Q+1)$, then we can say there are $C$ distinct roots with total multiplicity $9c$. If $r$ is a root of $(P+1,Q+1)$, then we can say there are $D$ distinct roots with total multiplicity $d$. Note that $A \leq a$ and similarly for the other pairs of variables. Now, note that $P$ will have exactly $20a+c$ roots with multiplicity (since each root contributes nine times) and $P+1$ will have exactly $20b+d$ roots with multiplicity (since each root contributes once). Likewise, $Q$ will have exactly $9a+b$ roots and $Q+1$ will have exactly $9c+d$ roots. It is well-known (say, by the product rule) that for any root of a polynomial $f$ with multiplicity $m$ is a root of $f'$ with multiplicity $m-1$. Thus, $P'$ has at least $20a+20b+c+d-A-B-C-D$ roots (with multiplicity). Thus, $$20a+20b+c+d-A-B-C-D \leq \deg P'=\deg P-1=20a+c-1 \implies 20b+d\leq A+B+C+D-1,$$and similarly $20a+c\leq A+B+C+D-1$ (from looking at $P+1$ instead of $P$), so $$20a+20b+c+d\leq 2A+2B+2C+2D-2 \leq 2a+2b+2c+2d-2 \implies 18a+18b+2 \leq c+d.$$On the other hand, this means that $$\deg Q=9a+b<18a+18b+2\leq c+d\leq 9c+d=\deg (Q+1),$$which is absurd, hence no such polynomials exist. $\blacksquare$

The answer is no. The key is the following lemma. Lemma. Let $\mathcal S(P)$ denote the set of roots, without multiplicity, of a polynomial $P$. Then $$|\mathcal S(P)| + |\mathcal S(P+1)| \geq \deg P + 1.$$ Proof. See here. (Indeed, the similarity to that problem has been pointed out previously in the thread.) $\blacksquare$ First, by comparing degrees, $\deg P = 21k$ and $\deg Q = 10k$ for some positive integer $k$. Now notice that $$|\mathcal S(P(x)^{10} + P(x)^9)| = |\mathcal S(P(x))| + |\mathcal S(P(x) + 1)| \geq 21k+1$$by the lemma. On the other hand, $$|\mathcal S(Q(x)^{21} + Q(x)^{20})| \leq \deg Q + \deg(Q+1) = 20k.$$This yields a contradiction. $\blacksquare$

i hate algebra i hate algebra i hate algebra i hate algebra why does it have to be calculus?? also missed the other P, P+1 question in DAW realpoly whoops

We will consider both the given equation $P^9(P+1) = Q^{20}(Q+1)$ and its derivative $P^8P'(10P+9) = Q^{19}Q'(21Q+20)$. The former implies that any root $r$ of $Q$ is a root of $P$ or $P+1$. In particular, $Q$ does not share any roots with $10P+9$, so $10P+9\mid Q'(21Q+20)$. However this is absurd since $\text{deg}P = \frac{21}{10}\text{deg}Q$, so no such polynomials exist. Edit: milestone post or something.

We have that $\deg(Q(x))=10n$ and $\deg (P(x))=21n$ for some $n$. Clearly we have that all the roots of $P(x)$ and all the roots of $P(x)+1$ are also roots of either $Q(x)$ or $Q(x)+1$. Let $P$ be the set of distinct roots of $P(x)$, $P_1$ be the set of distinct roots of $P(x)+1$ and define $Q$ and $Q_1$ similarly. We have that $\vert P \cup P_1 \vert \geq 21n+1$ by Putnam 1956 B7, thus we also get that as $\deg(Q(x))=10n$, $\vert Q \cup Q_1 \vert \leq 20n$, so we have a contradiction and thus no such polynomials exist.