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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
About old Inequality
perfect_square   0
4 minutes ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
perfect_square
4 minutes ago
0 replies
inquality
karasuno   1
N 27 minutes ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
1 viewing
karasuno
an hour ago
sqing
27 minutes ago
Number Theory
karasuno   0
an hour ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
an hour ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N an hour ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
an hour ago
No more topics!
Co-prime polynomials
hajimbrak   16
N Jan 18, 2025 by alexanderhamilton124
Source: Indian TST Day 1 Problem 1
Find all polynomials $f(x)$ with integer coefficients such that $f(n)$ and $f(2^{n})$ are co-prime for all natural numbers $n$.
16 replies
hajimbrak
Jul 11, 2014
alexanderhamilton124
Jan 18, 2025
Co-prime polynomials
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G H BBookmark kLocked kLocked NReply
Source: Indian TST Day 1 Problem 1
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hajimbrak
209 posts
#1 • 8 Y
Y by anantmudgal09, TheOneYouWant, doxuanlong15052000, megarnie, Master_of_Aops, Adventure10, CRT_07, and 1 other user
Find all polynomials $f(x)$ with integer coefficients such that $f(n)$ and $f(2^{n})$ are co-prime for all natural numbers $n$.
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joybangla
836 posts
#2 • 13 Y
Y by debjit1999, mbrc, quangminhltv99, biomathematics, doxuanlong15052000, Math-Ninja, Adventure10, Mango247, thepassionatepotato, CRT_07, and 3 other users
We will show $f\equiv 1$ and $f\equiv -1$ work.
Lemma :
$f(2^k)=1$ or $-1$ each $k$.
Proof :
Suppose not. Then take a prime $p$ dividing $f(2^k)$ for some $k$. Now see that $p\mid f(s_r)$ where $s_r=2^k+rp$ for any integer $r$. Now, consider $f(2^{s_r})$. We try to find an $r$ such that $p\mid 2^{s_r}-2^k$. Then we would have $p\mid 2^{s_r}-2^k\mid f(2^{s_r})-f(2^k)\implies p\mid f(2^{s_r})$ hence $\gcd(f(s_r),f(2^{s_r}))>1$ which serves our contradiction. It remains to find such an $r$. If $p=2$ then it is trivial. Suppose $p>2$. We need to find $r$ such that
\[ 2^{s_r}\equiv 2^k\pmod p\implies 2^{2^k-k+rp}\equiv 1\pmod p\] but since $p$ is odd hence $\{2^n\mod p\}$ is periodic. Hence such an $r$ exists. Hence we are done.
Now we have that for each $n,\; f(2^n)$ is $1$ or $-1$. Suppose there exist $k,\ell$ with $f(2^k)=1,f(2^\ell)=-1$. But then
\[ 2^k-2^\ell \mid 2 \] which is a contradiction for sufficiently large $k,\ell$. Then for big $k$ we have $f(2^k)=1$ or $-1$ throughout. Then $f(x)=1$ has infinitely many solutions, and similarly for $f(x)=-1$ and hence $f$ is identically equal to $1$ or $-1$. Hence our claim follows.
In the end, our solutions are :
\[ \boxed{f\equiv 1,f\equiv -1}\]
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AnonymousBunny
339 posts
#3 • 3 Y
Y by Adventure10, Mango247, CRT_07
Alternative finish (I believe this is how XmL did it too): The lemma in the post above shows that at least one of the polynomials $f(x)-1$ and $f(x)+1$ has infinitely many roots, so one of them must be identically equal to zero, so $f(x)$ is either always 1 or always -1.
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anantmudgal09
1979 posts
#4 • 2 Y
Y by Adventure10, CRT_07
I am posting in a hurry so if there is anything wrong please inform me. Thank you.

Here:
Let us observe that $f \equiv 1$ and $f\equiv -1$ are solutions and no other constant polynomial works.
Now assume that
$degP \ge 1$.
Now let $k_0$ be a natural number such that $\forall k \ge k_0$, $f(k)$ has absolute value greater than $1$.

Let $p$ be a prime such that $p \mid f(2^k)$ and let $p>2$ (Such a prime $p$ exists by taking $k$ to be larger than $v_2(f(0))$ or in other words $k$ to be sufficiently large.)
.
Then set $n= 2^k +p(l(p-1)+k-2^k)$ for sufficiently large $l,k \in \mathbb{N}$.
Now, $f(n) \equiv f(2^k) \equiv 0 \bmod p$
and $f(2^n) \equiv f(2^k) \equiv 0 \bmod p$ because of our choice of $n$.
This clearly leads to a contradiction to our hypothesis.
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PRO2000
239 posts
#6 • 3 Y
Y by Adventure10, Mango247, CRT_07
Assume that $f$ is nonconstant . Consider $(f(2^i))_{i \geq 1}$ . For sufficiently large $k$ , by our assumption that $f$ is nonconstant , we may choose a prime divisor $p$ so that $p|f(2^k)$ and $p \geq 2$ . Choose $n \equiv 2^k \text{( mod p )}$ and $n \equiv k \text{( mod (p-1) )}$.

Then , $f(2^n) \equiv f(n) \equiv 0 \text{( mod p )}$ . This is a contradiction , so $f$ is constant and it is easy to conclude that $f \equiv +1 $ or $f \equiv -1$ .
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ayan.nmath
643 posts
#7 • 4 Y
Y by paragdey01, Adventure10, Mango247, CRT_07
Indian TST Day 1 Problem 1 wrote:
Find all polynomials $f(x)$ with integer coefficients such that $f(n)$ and $f(2^{n})$ are co-prime for all natural numbers $n$.
Solution
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aops29
452 posts
#8 • 4 Y
Y by Alireza_Amiri, Adventure10, Mango247, CRT_07
Solution
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Alireza_Amiri
82 posts
#9 • 4 Y
Y by Mango247, Mango247, Mango247, CRT_07
Very nice solution @aops29 but in this part:
aops29 wrote:
Take \(l\) to be an integer such that \(l\equiv m - 2^m \pmod{p}\).
I think it should be $\pmod{p-1}$
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TFIRSTMGMEDALIST
162 posts
#10 • 1 Y
Y by CRT_07
how did he prove it
Attachments:
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Olympikus
87 posts
#11 • 1 Y
Y by CRT_07
Suppose there exists $p$ odd prime with $p\mid f(2^{n_0})$. Then, using the chinese remainder theorem, take
\[n \equiv n_0 \pmod{p-1}\]\[n\equiv 2^{n_0} \pmod p.\]Since if $a\equiv b \pmod n$, $f(a)\equiv f(b) \pmod n$,
\[f(n) \equiv f(2^{n_0}) \equiv f(2^n) \pmod p,\]a contradiction.
Hence, for each $n,$ $\pm f(2^n)$ is a power of two. Then, if it is even for some $n, \ f(0)$ is even, so $2\mid \gcd (f(2),f(4))$.
Therefore, $f(2^n) = \pm 1$ for all $n$, so either $f+1$ or $f-1$ has infinitely many zeros. This implies that $f\equiv 1$ or $f\equiv -1$, which both work.
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TFIRSTMGMEDALIST
162 posts
#12 • 1 Y
Y by CRT_07
Olympikus wrote:
Suppose there exists $p$ odd prime with $p\mid f(2^{n_0})$. Then, using the chinese remainder theorem, take
\[n \equiv n_0 \pmod{p-1}\]\[n\equiv 2^{n_0} \pmod p.\]Since if $a\equiv b \pmod n$, $f(a)\equiv f(b) \pmod n$,
\[f(n) \equiv f(2^{n_0}) \equiv f(2^n) \pmod p,\]a contradiction.
Hence, for each $n,$ $\pm f(2^n)$ is a power of two. Then, if it is even for some $n, \ f(0)$ is even, so $2\mid \gcd (f(2),f(4))$.
Therefore, $f(2^n) = \pm 1$ for all $n$, so either $f+1$ or $f-1$ has infinitely many zeros. This implies that $f\equiv 1$ or $f\equiv -1$, which both work.

In chinese remainder u can choose whatever you want ? Ur choice for $n_0$ and $2^n_0$
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TFIRSTMGMEDALIST
162 posts
#13 • 1 Y
Y by CRT_07
TFIRSTMGMEDALIST wrote:
how did he prove it

Help please
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CT17
1481 posts
#14 • 1 Y
Y by CRT_07
If $f$ is constant, then the only solutions are clearly $f\equiv -1$ and $f\equiv 1$.

Assume $f$ is nonconstant. Let $p$ be a prime dividing $f(2^k)$ for some positive integer $k$. Then $f(n)$ and $f(2^n)$ are both divisible by $p$ for any $n\equiv kp - 2^k(p-1)\pmod{p^2 - p}$ so we have a contradiction, as desired.
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L13832
245 posts
#15 • 1 Y
Y by CRT_07
Standard
solution
This post has been edited 1 time. Last edited by L13832, Jan 10, 2025, 3:28 AM
Reason: Typo
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onyqz
195 posts
#16 • 1 Y
Y by CRT_07
solution
motivation
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lelouchvigeo
172 posts
#17 • 1 Y
Y by CRT_07
let $p$ be a prime such that $p \mid f(2^k)$ .Then we have that $p \mid f(2^k+ cp)$, since $ 2^k \equiv2^k + cp$$  (mod p)$
We can choose $c$ such that $2^{2^k + cp} \equiv 2^ k$, $c \equiv k - 2^k$$ (mod p-1)$ Then we have $p \mid f(2^{2^k + cp} ) - f(2^k) \implies f \mid f(2^{2^k + cp} )$ . A contradiction.
Which implies $f(2^k) = \pm 1$ for every $k$ which is clearly not possible.
So $f$ is a constant polynomial and the solutions are $\boxed{f\equiv \pm 1}$.
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alexanderhamilton124
374 posts
#19 • 2 Y
Y by TensorGuy666, CRT_07
If it isn't $1$ or $-1$, consider a prime $p$ such that $p \mid f(2^x)$, for some $x$. Then, take $n \equiv 2^x \mod{p}$, $n \equiv k \mod{p - 1}$, which exists by CRT, a contradiction.
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