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jlacosta   0
May 1, 2025
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jlacosta
May 1, 2025
0 replies
Preparing for Putnam level entrance examinations
Cats_on_a_computer   1
N an hour ago by Miquel-point
Non American high schooler in the equivalent of grade 12 here. Where I live, two the best undergraduates program in the country accepts students based on a common entrance exam. The first half of the exam is “screening”, with 4 options being presented per question, each of which one has to assign a True or False. This first half is about the difficulty of an average AIME, or JEE Adv paper, and it is a requirement for any candidate to achieve at least 24/40 on this half for the examiners to even consider grading the second part. The second part consists of long form questions, and I have, no joke, seen them literally rip off, verbatim, Putnam A6s. Some of the problems are generally standard textbook problems in certain undergrad courses but obviously that doesn’t translate it to being doable for high school students. I’ve effectively got to prepare for a slightly nerfed Putnam, if you will, and so I’ve been looking for resources (not just problems) for Putnam level questions. Does anyone have any suggestions?
1 reply
Cats_on_a_computer
2 hours ago
Miquel-point
an hour ago
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   8
N 4 hours ago by Aiden-1089
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
8 replies
Silver08
Today at 2:26 AM
Aiden-1089
4 hours ago
f(x+1)-f(x)=f'(x+1/2) implies f(x)=ax^2 +bx+c?
tom-nowy   1
N 4 hours ago by ddot1
Source: https://artofproblemsolving.com/community/c4t157249f4h1288200
Is this true?

$f: \mathbb{R} \to \mathbb{R}$ is differentiable and for all $x \in \mathbb{R}, \; f(x+1)-f(x)=f'\left(x+\frac{1}{2}\right)$
$\Longrightarrow f(x)=ax^2 +bx+c$.
1 reply
tom-nowy
Today at 2:47 AM
ddot1
4 hours ago
Integration Bee Kaizo
Calcul8er   57
N Today at 2:00 AM by Silver08
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
57 replies
Calcul8er
Mar 2, 2025
Silver08
Today at 2:00 AM
No more topics!
Putnam 2000 A6
ahaanomegas   15
N Apr 6, 2025 by Levieee
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.
15 replies
ahaanomegas
Sep 6, 2011
Levieee
Apr 6, 2025
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ahaanomegas
6294 posts
#1 • 3 Y
Y by Rounak_iitr, Adventure10, Mango247
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.
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Kent Merryfield
18574 posts
#2 • 9 Y
Y by FinalSix, skrublord420, Centralorbit, math_and_me, Arkajit_Ganguly, Adventure10, Mango247, and 2 other users
If $a_1 = 0,$ then $a_n = 0\  \forall\, n$ and we are done. In all that follows, assume $a_1 \ne 0.$

Let $N$ be the smallest positive integer for which $a_N = 0.$ That there is such an N comes from the hypothesis of the problem that the sequence eventually reaches $0$ and from the well-ordering of the natural numbers. By our assumption, $N \ge 2.$

If for any $k < m, a_k = a_m,$ then for all $n > m,$ if we find that $s, k \le s < m,$ for which $n \equiv s \pmod {(m - k)},$ then $a_n = a_s.$ In other words, from that point on, the sequence $a_n$ is periodic of period $m - k.$ This follows from the fact that each $a_n$ depends only on its predecessor and an induction.

It now follows that we cannot have $a_k = a_m$ for any $0 < k < m < N,$ since if that did occur, the sequence would repeat only those values of $a_n$ for $k \le n < m,$ none of which are zero.

We observe that if $f$ is an integer-coefficient polynomial and $x$ and $y$ are integers, then $(x - y)\, |\, (f(x) - f(y)).$ This follows by induction and linearity from the fact that for each integer power $k, (x - y)\, |\, (x^k - y^k).$ This is clear for $k = 0$ and $1,$ and for larger $k,$ $(x_k - y_k) = (x - y)(x^{k-1} + x^{k-2}y + \cdots + y^{k-1}).$ Then, since $a_{n+2} - a_{n+1} = f(a_{n+1}) - f(a_n),$ we have that for all $n \ge 0,$ $(a_{n+1} - a_n)\, |\, (a_{n+2} - a_{n+1}).$ Each difference divides the next difference. None of these differences are zero (by the previous $a_k \ne a_m$ argument) but since the sequences are periodic, the differences are periodic. That means that no difference can ever be bigger (in absolute value) than any other difference: they are all, except possibly for sign, the same. Since the first difference is $a_1,$ all differences are $\pm a_1.$ That means that the sequence $a_n$ “walks” one step at a time, backwards or forwards, over the set of integer multiples of $a_1.$ This walking sequence starts at $0$ and must return to $0.$ However, from the $a_k \ne a_m$ argument, this sequence can never retrace any steps on the way from 0 to 0, and that means that it can never get two steps away from 0. Its first step is to $a_1;$ its next step cannot be to $2a_1$ so it must be back to $0.$ Hence, $a_2 = 0,$ which is what we were trying to prove.
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JSGandora
4216 posts
#3 • 3 Y
Y by Euler1728, tuanngoc298, Adventure10
Notice that $a_n=f^n(0)$ and $a_0=0$. Since $f^{m}(0)=f(f^{m-1}(0))=0$, we must have $f^{m-1}(0)$ to be a root of $f(x)$. Call this root $r$. Also call the constant term of $f$, $c_0$ and notice that $f(0)=c_0$.

There are two cases, where $c_0=0$ and where $c_0\neq 0$. If $c_0=0$ then $a_1=f(0)=c_0=0$ and we are done. If $c_0\neq 0$, we proceed as follows.

Notice that $c_0 | f(kc_0)$ where $k\in \mathbb{Z}$ since $f$ has integer coefficients. Then we have
\[c_0 | f(0)=c_0\]
so
\[
c_0 | f(0) | f(f(0)) | \cdots | f^{m-1}(0) = r
\]
Hence $c_0|r$. However, since $r$ is clearly an integer, by the rational root theorem, $r|c_0$ and thus we have $c_0=r$. Therefore, $a_2 = f(f(0))=f(c_0)=f(r)=0$.

Therefore, either $a_1=0$ or $a_2=0$ as desired.
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e_plus_pi
756 posts
#4 • 3 Y
Y by GreenKeeper, Adventure10, Mango247
Nice Problem.
Here is my solution:
$\longrightarrow$ Since $f(x)$ is an integer polynomial, we use the fact that $ m-n| f(m)-f(n) \forall m,n \in \mathbb{Z}$.

Let $ b_n =^{DEF} a_{n+1}-a_n$ . Then, $b_n | b_{n+1} \forall n$.

On the other hand, we are given that $a_0 = a_m = 0$ so $a_n$ is cyclic after m terms and so $b_0 = b_m$.

If $b_0 =0 $, then $a_i = a_j \forall i,j\in (0,1,2,\dots,m)$ and we are done. Otherwise $|b_0| = |b_1| =\dots $

So $ b_n = b_0$ or $b_n = - b_0$ and from here the conclusion follows easily.
This post has been edited 1 time. Last edited by e_plus_pi, Apr 26, 2018, 2:50 PM
Reason: Spacing
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Luta_Tnana
18 posts
#5 • 1 Y
Y by Adventure10
Thanks a lot
This post has been edited 1 time. Last edited by Luta_Tnana, Jul 1, 2019, 8:31 PM
Reason: I understood it later
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Luta_Tnana
18 posts
#6 • 1 Y
Y by Adventure10
Oh I get ya ! Sorry
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Ramanujan_math
375 posts
#7 • 1 Y
Y by Adventure10
ahaanomegas wrote:
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.

This is a question of ISI 2019
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Luta_Tnana
18 posts
#8 • 1 Y
Y by Adventure10
Ya it is !!
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Raunii
28 posts
#11
Y by
JSGandora wrote:
Notice that $a_n=f^n(0)$ and $a_0=0$. Since $f^{m}(0)=f(f^{m-1}(0))=0$, we must have $f^{m-1}(0)$ to be a root of $f(x)$. Call this root $r$. Also call the constant term of $f$, $c_0$ and notice that $f(0)=c_0$.

There are two cases, where $c_0=0$ and where $c_0\neq 0$. If $c_0=0$ then $a_1=f(0)=c_0=0$ and we are done. If $c_0\neq 0$, we proceed as follows.

Notice that $c_0 | f(kc_0)$ where $k\in \mathbb{Z}$ since $f$ has integer coefficients. Then we have
\[c_0 | f(0)=c_0\]so
\[
c_0 | f(0) | f(f(0)) | \cdots | f^{m-1}(0) = r
\]Hence $c_0|r$. However, since $r$ is clearly an integer, by the rational root theorem, $r|c_0$ and thus we have $c_0=r$. Therefore, $a_2 = f(f(0))=f(c_0)=f(r)=0$.

Therefore, either $a_1=0$ or $a_2=0$ as desired.

it may happen that $c=-r$ then what? wont we have to take cases
This post has been edited 1 time. Last edited by Raunii, Mar 20, 2020, 4:03 PM
Reason: ,
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peatlo17
77 posts
#12
Y by
I agree with @above. I tried the same approach as @JSGandora with the rational root theorem, but was eventually stuck on the case $c = -r$.

I believe that the only way to resolve this is through the the differences / integer divisibility approach.
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lifeismathematics
1188 posts
#14
Y by
define $d_{i}=a_{i+1}-a_{i}$

Now since $f(x) \in \mathbb{Z}[x]$ , we have $a_{i+1}-a_{i}|f(a_{i+1})-f(a_{i}) =a_{i+2}-a_{i+1}=d_{i+1}$

so we have $d_{i}|d_{i+1}$ $\forall$ $i \geqslant 0$

now we also have $a_{m}=0$ for some $m \in \mathbb{Z}^{+}$ , so $a_{m}=a_{0}$ , and $a_{m+1}=a_{1}$ , hence we have $d_{m}=d_{0}$

hence we have $|d_{0}|=|d_{1}|=|d_{2}|=\cdots =|d_{i}|=\Omega$ for $i \geqslant 0$

Now:

  • if $\Omega=0$

    so we have $a_{1}=0$.
  • if $\Omega \neq 0$

    so we have that the set $\{ a_{1} , a_{2} , \cdots , a_{m}\} \subseteq \mathbb{Z}^{+}_{0}$ , hence by well ordering principle this set has a minimum. say $a_{\ell}$ , so $a_{\ell-1}>a_{\ell}$ , $a_{\ell+1}>a_{\ell}$ , so we have $d_{\ell-1}=-d_{\ell} \implies a_{\ell-1}=a_{\ell+1}$ , now we can take $f$ over until we get $a_{2}=a_{0}=0$ , or if not then we have $f(a_{\ell-2})=f(a_{\ell})\implies a_{\ell-2}=a_{\ell}$ so we do this until we get $a_{2}=a_{0}=0$

Hence either $a_{1}=0$ or $a_{2}=0.$ $\blacksquare$.
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chakrabortyahan
381 posts
#15
Y by
Lol isi 2019 problem...(not nearly a A6 though) in 2nd page there is a typo it would be $f(d)=2d,f(2d) = d$...and there was a third page with the line "Hence $a_2 = 0$ if $a_1\neq 0 $"
$\blacksquare\smiley$
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sanyalarnab
933 posts
#16
Y by
Storage purposes only... one of my most favorite problems :-D
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PRMOisTheHardestExam
409 posts
#17
Y by
lemma in imo 2006 p5
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Aiden-1089
285 posts
#18
Y by
We prove the following generalisation:
If a polynomial $f(x) \in \mathbb{Z}[x]$ has a degree $n$ fixed point $a$, then $n=1$ or $2$.

Assume FTSOC that $n>2$.
Note that for all integers $x \neq y$, we have $x-y \mid f(x)-f(y)$.
Let $x_i=f^i(a)-f^{i-1}(a)$ for all positive integers $i$, so $x_{n+i}=x_{i}$. Then we have $x_2 \mid x_3 \mid \cdots \mid x_{n+2} = x_2$, from which we see that $\left| x_i \right|$ is a nonzero constant for all positive integers $i$.
Note that $x_1=x_2$ since $n>2$, and $\sum_{i=1}^n x_i =0$. Since it is not possible that $x_i$ is constant, let $k$ be the smallest positive integer such that $x_{k+2}=-x_1$.
Then we have $f^{i}(a) = ix_1 + a$ for all $1 \leq i \leq k+1$ and $f^{k+2}(a)= kx_1+a = f^k(a)$.
It follows that $f^{k+m}(a) = \begin{cases} f^k(a)=kx_1+a & \text{if } m \text{ is even} \\ f^{k+1}(a)=(k+1)x_1+a &\text{if } m \text{ is odd} \end{cases}$ for all nonnegative integers $m$, so $f^m(a) \neq a$ for all positive integers $m$, which contradicts the fact that $f^n(a)=a$.
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Levieee
221 posts
#19
Y by
Define:= $b_{n}=a_{n+1}-a_{n}$
Lemma 1: if $f(x) \in \mathbb{Z}[x]$ then $x-y \mid f(x)-f(y)$ $\forall x,y \in \mathbb{Z}$

Coming back to the problem,
$f(a_{m})=a_{m+1}$
$f(0)=a_{m+1}$
we know $f(a_{0})=a_{1}$
$f(0)=a_{1}$
$a_{1}=a_{m+1}$
Similarly by induction we can show that $a_{m}=a_{m+k}$ $\forall k \in \mathbb{N}_{0}$
$\therefore$ $(a_{n})_{n\geq 1}$ is a cyclic sequence in $m$ terms

Since we know that the sequence $(a_{n})_{n \geq 1}$ is an integer sequence therefore Lemma 1 is applicable to all the terms in the sequence

Now,
$a_{n+1}-a_{n} \mid f(a_{n+1})-f(a_{n})$
or,$b_{n} \mid b_{n+1}$

from the fact that $a_{n}$ is a cyclic sequence in $m$ terms notice that $b_{m}$ is cyclic in $m$ terms,

$\therefore$ |$b_{0}$|=|$b_{m}$|

We know that $b_{n} \mid b_{n+1}$ and $(b_{n})_{n\geq 1}$ is a cyclic sequence in $m$ terms

therefore $|b_0| = |b_1| =\dots $

Case 1:

$b_0 =0 $

Then $a_i = a_j \forall i,j\in (0,1,2,\dots,m)$ and we are done


Case 2:

$|b_0| = |b_1| =\dots $ $= \alpha$

$ b_n =  b_0$ or $b_n = - b_0$

$a_{n+1}-a_{n}=a_{1}$ or $a_{n+1}-a_{n}=-a_{1}$
Hence, the conclusion follows $\mathbb{Q.E.D}$ $\blacksquare$
:whistling:
This post has been edited 2 times. Last edited by Levieee, Apr 6, 2025, 1:37 PM
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