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k a 2024 AMC 10/12B Math Jam!
LauraZed   0
Wednesday at 5:10 PM
Want to discuss the AMC 10/12B with AoPS instructors? Join us for the 2024 AMC 10B/12B Math Jam tonight, November 13, at at 7:30pm ET / 4:30pm PT, where we will discuss some of the most interesting problems from each test!

You can learn how to attend Math Jams here:
https://artofproblemsolving.com/school/mathjams

Finally, the Contests & Programs forum has also reopened for discussion of these exams, and you can find the official discussion threads here:
https://artofproblemsolving.com/community/c5h3442595_2024_amc_10b_discussion_thread
https://artofproblemsolving.com/community/c5h3442596_2024_amc_12b_discussion_thread
0 replies
LauraZed
Wednesday at 5:10 PM
0 replies
k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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0 replies
jlacosta
Nov 1, 2024
0 replies
arctan inequality
tohill   2
N 7 minutes ago by mathyperson
Source: own
prove this inequality :
2arctan[(y-x)/2]≥arctany-arctanx ,y≥x
2 replies
+1 w
tohill
Yesterday at 3:14 PM
mathyperson
7 minutes ago
Convexity proof
MetaphysicalWukong   4
N an hour ago by tobiSALT
Source: Zhoujin Xia, Dunbo Yang
if f is twice differentiable on I, then f is convex if and only if f''(x)≥0 ∀ x∈I.
4 replies
MetaphysicalWukong
Nov 13, 2024
tobiSALT
an hour ago
conjugate analysis
ironball   2
N an hour ago by tobiSALT
Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
2 replies
ironball
Nov 11, 2024
tobiSALT
an hour ago
Theorem in algebra concerning equalities
τρικλινο   33
N 3 hours ago by Amkan2022
if a,b are greater or equal to zero then prove:
$b^2=a^2$ implies that b=a
33 replies
τρικλινο
Nov 9, 2024
Amkan2022
3 hours ago
No more topics!
Putnam 2010 B4
Kent Merryfield   13
N Yesterday at 2:37 PM by Ihatecombin
Find all pairs of polynomials $p(x)$ and $q(x)$ with real coefficients for which
\[p(x)q(x+1)-p(x+1)q(x)=1.\]
13 replies
Kent Merryfield
Dec 6, 2010
Ihatecombin
Yesterday at 2:37 PM
Putnam 2010 B4
G H J
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Kent Merryfield
18574 posts
#1 • 3 Y
Y by Rounak_iitr, Adventure10, Mango247
Find all pairs of polynomials $p(x)$ and $q(x)$ with real coefficients for which
\[p(x)q(x+1)-p(x+1)q(x)=1.\]
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yjneb
62 posts
#2 • 3 Y
Y by Adventure10, Rounak_iitr, Mango247
Plug in x-1 and manipulate two eqns to get p(x)/(p(x-1)+p(x+1))= same expression for q. But p and q clearly rel. prime by given eqn, so reduced numerator must be 1, so p(x) divides p(x-1)+p(x+1), but looking at leading coef. we see 2p(x)=p(x-1)+p(x+1). We can induct and find p(n)=p(0)+n(p(0)-p(-1)), giving infinitely many agreements with p(x)=p(0)+x(p(0)-p(-1)), and any two polynomials that agree infinitely often are the same. Thus p must be linear or constant. Similarly, q must also be constant or linear. Writing p(x)=ax+b, q(x)=cx+d and plugging in gives that the given condition is equivalent to bc-ad=1, (or something like that).
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jmerry
12096 posts
#3 • 3 Y
Y by Adventure10, Rounak_iitr, Mango247
The only such pairs are the pairs of linear and/or constant polynomials $p(x)=a_1x+b_1$, $q(x)=a_2x+b_2$ with $a_2b_1-a_1b_2=1$.

Define $p(x)*q(x)=p(x)q(x+1)-p(x+1)q(x)$. $*$ is a bilinear, alternating operator from pairs of polynomials to pairs of polynomials, much like the cross product.
By the bilinearity, we can find $p*q$ by taking a linear combination of terms of the form $x^n*x^m$. The exact values of these are complicated, but $x^n*x^m$ is a polynomial of degree $n+m-1$ (with leading term $(m-n)x^{n+m-1}$) whenever $m\neq n$. Therefore, if $p$ and $q$ have distinct degrees $m$ and $n$, $p*q$ has degree $m+n-1$.
We still have to consider the case in which $p$ and $q$ have the same degree $m$. Then we can write $p=a\cdot q+r$, for some constant $a$ and some polynomial $r$ of degree $n<m$. By the bilinearity and alternating properties, $p*q=(a\cdot q+r)*q=a(q*q)+r*q=r*q$, which has degree $m+n-1$.
For this $m+n-1$ to be zero, $m$ and $n$ must be zero and $1$ in some order; either we started with a linear polynomial and a constant, or we started with two linear polynomials and reduced from there. Calculating directly, $(a_1x+b_1)*(a_2x+b_2)=a_2b_1-a_1b_2$.
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fedja
6920 posts
#4 • 8 Y
Y by PRO2000, Adventure10, Rounak_iitr, Mango247, Ihatecombin, and 3 other users
Here is a solution with "analytic" flavor:

Let $r(x)=\frac{p(x)}{q(x)}$. Then $r(x+1)=r(x)-\frac 1{q(x)q(x+1)}$. If $q$ has degree $n>1$, we conclude that the limit $a=\lim_{x\in\mathbb Z,x\to\infty}r(x)$ exists and $r(x)=a+O(x^{1-2n})$. On the other hand, $r(x)-a$ is a rational function that is not identically $0$ and has denominator of degree $n$, so it can decay as $x^{-n}$ at best, which leads to a contradiction since $2n-1>n$. Similarly, the degree of $p$ is at most $1$. The rest is just trivial computation.
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Zhero
2043 posts
#5 • 6 Y
Y by mutasimmim12, arandomperson123, lukamacho2001, Rounak_iitr, Adventure10, Kingsbane2139
Lemma: Suppose $r(x)$ is a polynomial such that $r(x+1)$ divides $r(x) + r(x+2)$. Then $r$ must be linear.
Proof: $r(x+1)$ must divide $r(x) + r(x+2) - 2r(x+1)$, which has degree less than that of $r$ (as the leading coefficients of the polynomials cancel.) Hence, $r(x) + r(x+2) - 2r(x+1)$ must be identically zero, so $r(x+2) = 2r(x+1) - r(x)$ must hold for all $x$. With this, it can easily be shown by induction that $r(n) = r(0) + n(r(1) - r(0))$ for all positive integers $n$. Since $r$ agrees with the linear function $r(0) + n(r(1) - r(0))$ for infinitely many $n$, $r$ must be linear. $\blacksquare$

We have $ p(x)q(x+1) - p(x+1) q(x) = 1$ and $p(x+1)q(x+2) - p(x+2)q(x+1) = 1$. Subtracting these two equations and rearranging yields $p(x+1)(q(x) + q(x+2)) = -q(x+1)(p(x) + p(x+2))$.

From $p(x)q(x+1) - p(x+1)q(x) = 1$, it is clear that $p$ and $q$ have no common factors. Hence, by Euclid's lemma on polynomials, $p(x+1)$ must divide $p(x) + p(x+2)$ and $q(x+1)$ must divide $q(x) + q(x+2)$, so by our lemma both $p$ and $q$ must be linear. Substituting $p(x) = ax + b$ and $q(x) = cx + d$ into the equation, we find that $1 = (ax + b)(cx + c + d) - (ax + a + b)(cx + d) = bc - ad$, so the set of polynomials $(p,q)$ satisfying the conditions of the problem are exactly the linear polynomials $(ax + b, cx + d)$ in which $bc - ad = 1$.
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Altheman
6199 posts
#6 • 2 Y
Y by Adventure10, Rounak_iitr
$1=p(x)q(x+1)-p(x+1)q(x)=p(x)q(x+1)-p(x)q(x)-p(x+1)q(x)+p(x)q(x)$
$=p(x)[ q(x+1)-q(x)]-q(x)[p(x+1)-p(x)]$

Assume that $p(x)=a_n x^n+...$, $q(x)=b_m x^m+...$. Then looking at the lead coefficient of that, we have

$a_n (m b_m)-b_m (na_n)=(m-n)a_nb_mx^{m+n-1}$

So If $m\ne n$, then $m+n-1=0$ and the problem is trivial. If the polynomials are the same degree, notice that

$[p,q]=1$ implies $[p-aq,p]=1$ also where $[p,q]$ denotes that combination of the polynomials. Thus starting with any pair, if they have the same degree, then we just subtract as such to get two poly. that have a different degree. I had also noticed that this function was bilinear but it didn't end up being too significant (although it let you handle the equal degrees case easily).
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blankspace
60 posts
#7 • 2 Y
Y by Adventure10, Mango247
random bump but wow fedja's solution is very nice!
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Vexation
49 posts
#8 • 3 Y
Y by Adventure10, Rounak_iitr, Mango247
Somewhat similar to fedja's...

Let $r(x) = \frac{p(x)}{q(x)}$. If $deg(p) > deg(q)$,
\[ r(x+1)-r(x) = \frac{1}{p(x)p(x+1)} \]the left side diverges from $O( r(x) ) \ge x$, while the right side converges to $0$. Hence, $deg(p)=deg(q)$.

By mean value theorem, for $c \in (x, x+1)$, \[ r(x+1)-r(x) = \frac{q'(c)p(c)-q(c)p'(c)}{p(c)^2} \]Let $s(x) = q'(x)p(x) - q(x)p'(x)$.
Since $s(x)$ is also a polynomial, for $|x|$ big enough, \[|s(c)| \ge \min \{|s(x)|, |s(x+1)| \}.\]Therefore,
\[ | {\frac{1}{p(x)(p(x+1)}} | = |r(x+1)-r(x)| = \frac{|s(c)|}{p(c)^2} \ge \min \{ \frac{|s(x)|}{p(c)^2} , \frac{|s(x+1)|}{p(c)^2}\}.\]\[ \frac{p(c)^2}{|p(x)p(x+1)|} \ge \min \{|s(x)|, |s(x+1)|. \} \]When $x$ diverges to $\infty$ and $-\infty$, the left side converges to $1$. Therefore, $s(x) = k$ for a constant $k$.

Let $r(x)$ be a polynomial whose degree is smaller than $q(x)$ and $q(x)= \alpha p(x)+r(x)$ for a constant $\alpha$.

\[q'(x)p(x) - q(x)p'(x) = (\alpha p(x) + r(x))'p(x) - (\alpha p(x) + r(x))p'(x) = r'(x)p(x) - r(x)p'(x) = k.\]
When $p(x)$ is the degree of $n$, the highest degree of $r(x)p'(x) - r(x)p'(x)$ is $x^{2n-2}$. $n=1$.
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anantmudgal09
1977 posts
#9 • 5 Y
Y by sa2001, e_plus_pi, everythingpi3141592, Rounak_iitr, Adventure10
We can assume that degrees of both are at least $1$. Note that $p, q$ have no common factor over $\mathbb{R}[X]$ from the given condition. Replacing $x \rightarrow x-1$ and subtracting we get $$p(x) \cdot \left(q(x+1)+q(x-1)\right)=q(x)\cdot \left(p(x+1)+p(x-1)\right),$$for all $x$. It follows that $p(x)$ divides $p(x+1)+p(x-1)$. Clearly, their quotient is a constant and comparing the leading coefficients gives $p(x+1)=2p(x)-p(x-1).$ Solving as a recurrence we get $$p(n+1)=n\cdot (p(1)-p(0))+p(0)$$for all integers $n>0$. As $p$ agrees with a linear function infinitely often; it is a linear function. Same argument works for $q$.

In conclusion, the only satisfactory polynomials are $(ax+b, cx+d)$ with $bc-ad=1$. These satisfy the equation so we are done.
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Rounak_iitr
447 posts
#11
Y by
$\color{red}\textbf{Claim:-}$ The pairs of polynomial $P(x)$ and $Q(x)$ works that is $P(x)=ax+b$ and $Q(x)=cx+d$ where $a,b,c,d \in \mathbb{Z}.$
$\color{blue}\textbf{Proof:-}$ Assume $P(x)=ax+b$ and $Q(x)=cx+d$ where $a,b,c,d \in \mathbb{Z}.$ satisfying the condition we get, $$P(x)Q(x+1)-P(x+1)Q(x)=1$$$$(ax+b)(cx+c+d)-(cx+d)(ax+a+b)=1\implies\boxed{bc-ad=1.}$$Now Plugging $x=x-1$ we get, $$P(x)Q(x+1)-P(x+1)Q(x)=1$$$$P(x-1)Q(x)-P(x)Q(x-1)=1$$Now subtracting these two equations we get, $$P(x)[Q(x+1)-Q(x-1)]=Q(x)[P(x+1)-P(x-1)]$$From these we can say that $$P(x)|P(x+1)-P(x-1) \hspace{2mm} \text{and} \hspace{2mm} Q(x)|Q(x+1)-Q(x-1)$$Since $P(x)\nmid Q(x).$ Now let's Calculate some expressions $$P(x+1)+P(x-1)=2(ax+b)\implies 2P(x).$$Similarly we get, $$Q(x+1)+Q(x-1)=2(cx+d)\implies 2Q(x).$$Now Let's Call $f(x)$ be the difference of $P(x)$ and $P(x-1)$ we get, $$f(x)=P(x)-P(x-1)$$Plugging $x=x+1$ we get, $$f(x+1)=P(x+1)-P(x)\implies f(x)$$Similarly now plugging $x=x+2$ we get, $$f(x+2)=P(x+2)-P(x+1)\implies f(x+1)$$And the cycle goes on. Now Similarly Call $g(x)$ be the difference of $Q(x)$ and $Q(x-1)$ we get, $$g(x)=Q(x)-Q(x-1)$$Plugging $x=x+1$ we get, $$g(x+1)=Q(x+1)-Q(x)\implies g(x).$$Similarly now plugging $x=x+2$ we get, $$g(x+2)=Q(x+2)-Q(x+1)\implies g(x+1).$$Now Plugging $x=0$ in $f(x)$ we get, $$f(0)=P(0)-P(-1)\implies a.\implies P(0)=b.$$Now plugging $x=0$ in $g(x)$ we get, $$g(0)=Q(0)-Q(-1)\implies c.\implies Q(0)=d.$$Therefore we get, $$P(x)Q(x+1)-P(x+1)Q(x)=bc-ad=1.$$
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exopeng
21 posts
#12
Y by
Rounak_iitr wrote:
$\color{red}\textbf{Claim:-}$ The pairs of polynomial $P(x)$ and $Q(x)$ works that is $P(x)=ax+b$ and $Q(x)=cx+d$ where $a,b,c,d \in \mathbb{Z}.$
$\color{blue}\textbf{Proof:-}$ Assume $P(x)=ax+b$ and $Q(x)=cx+d$ where $a,b,c,d \in \mathbb{Z}.$ satisfying the condition we get, $$P(x)Q(x+1)-P(x+1)Q(x)=1$$$$(ax+b)(cx+c+d)-(cx+d)(ax+a+b)=1\implies\boxed{bc-ad=1.}$$Now Plugging $x=x-1$ we get, $$P(x)Q(x+1)-P(x+1)Q(x)=1$$$$P(x-1)Q(x)-P(x)Q(x-1)=1$$Now subtracting these two equations we get, $$P(x)[Q(x+1)-Q(x-1)]=Q(x)[P(x+1)-P(x-1)]$$From these we can say that $$P(x)|P(x+1)-P(x-1) \hspace{2mm} \text{and} \hspace{2mm} Q(x)|Q(x+1)-Q(x-1)$$Since $P(x)\nmid Q(x).$ Now let's Calculate some expressions $$P(x+1)+P(x-1)=2(ax+b)\implies 2P(x).$$Similarly we get, $$Q(x+1)+Q(x-1)=2(cx+d)\implies 2Q(x).$$Now Let's Call $f(x)$ be the difference of $P(x)$ and $P(x-1)$ we get, $$f(x)=P(x)-P(x-1)$$Plugging $x=x+1$ we get, $$f(x+1)=P(x+1)-P(x)\implies f(x)$$Similarly now plugging $x=x+2$ we get, $$f(x+2)=P(x+2)-P(x+1)\implies f(x+1)$$And the cycle goes on. Now Similarly Call $g(x)$ be the difference of $Q(x)$ and $Q(x-1)$ we get, $$g(x)=Q(x)-Q(x-1)$$Plugging $x=x+1$ we get, $$g(x+1)=Q(x+1)-Q(x)\implies g(x).$$Similarly now plugging $x=x+2$ we get, $$g(x+2)=Q(x+2)-Q(x+1)\implies g(x+1).$$Now Plugging $x=0$ in $f(x)$ we get, $$f(0)=P(0)-P(-1)\implies a.\implies P(0)=b.$$Now plugging $x=0$ in $g(x)$ we get, $$g(0)=Q(0)-Q(-1)\implies c.\implies Q(0)=d.$$Therefore we get, $$P(x)Q(x+1)-P(x+1)Q(x)=bc-ad=1.$$

How is this a proof of the problem?
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sanyalarnab
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:coolspeak:
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IAmTheHazard
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Obviously the equation holds as a polynomial identity. Let $A(x)=p(x)q(x+1)$ and $B(x)=p(x+1)q(x)$. Then $A$ and $B$ should have all coefficients the same except for the constant term, and should have the same degree $d$. Thus the first $d-1$ Newton sums of $A$ and $B$ should be the same.

Let $p_1,\ldots,p_n$ be the roots (with multiplicity) of $p$ and $q_1,\ldots,q_m$ be the roots of $q$. The above condition then implies that
$$\sum_{i=1}^n p_i^k+\sum_{i=1}^m (q_i-1)^k=\sum_{i=1}^n (p_i-1)^k+\sum_{i=1}^m q_i^k$$for all $1 \leq k \leq d-1$. Let $f_k(x)=x^k-(x-1)^k$ be a degree $k-1$ polynomial; the above sum rearranges to $\sum f_k(p_i)=\sum f_k(q_i)$ for all $1 \leq k \leq d-1$, so by induction we get $\sum p_i^k=\sum q_i^k$ for all $0 \leq k \leq d-2$. In particular from $k=0$ we obtain $m=n$, but now if $n \geq 2$ we have $d-2=2n-2 \geq n$, yet the first through $n$-th Newton sums of the roots of $p$ and $q$ are the same, so $p$ and $q$ would be the same up to a constant factor. However, this is absurd, so we obtain $n \leq 1$ and the rest is easy.
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Ihatecombin
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This is my attempted proof, I’m a bit suspicious of it as it seems to be very different than everyone else's, a verification would be nice.

We claim the only solutions are $p(x) = ax+b$ and $q(x) = cx+d$ where $ad - bc = 1$. It is easy to check that this in fact does fulfill the problem statement. Now we will show that no other solution exists. Denote deg($p(x)$) = $m$ and deg($q(x)$) = $n$. WLOG $m$ $\geq$ $n$, now we shall assume for the sake of contradiction that $m > 1$, this clearly also forces $n \geq 1$ as if $q(x)$ was a constant then $p(x)$ cannot possibly have degree larger than 1.

Notice that by Euclid's division lemma we have $p(x) = q(x)H(x) + R(x)$, where deg$(R(x))$ < deg($q(x)$). We also have that $R(x)$ is not zero as clearly $p(x)$ is relatively prime to $q(x)$, and we have $n \geq 1$. Now by plugging in our value of $p(x)$ we obtain
\begin{align*}
    [q(x)H(x) + R(x)][q(x+1)] - [q(x+1)H(x+1) + R(x+1)][q(x)] &= 1 \\
    q(x+1)q(x)[H(x)-H(x+1)] + R(x)q(x+1) - R(x+1)q(x) &= 1
\end{align*}Notice that deg($R(x)) < m$ and thus the L.H.S is clearly a degree $2m$ polynomial if $H(x)$ is not a constant, however the R.H.S is a degree $0$ polynomial meaning $m = 0$, which is a clear contradiction. Therefore $H(x)$ is a constant and thus $n = m$ and
\[R(x)q(x+1) - R(x+1)q(x) = 1\]However this is just the original problem with the added caveat that in the previous problem since $H(x)$ is a constant, we had $n = m$. However in this case, we must have deg($R(x)) < m$, we'll also have that deg($R(x)$) $\geq 1$ because if it were a constant we would clearly have that $q(x)$ is linear and thus that $p(x)$ is linear (because $n = m$) which is a contradiction. Therefore deg($rRx)$) $\geq 1$ and $m \geq 2$, which is an exact repeat of our initial conditions, whence we obtain our contradiction thus we are done.
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