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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by RMO 2006
sqing   0
a minute ago
Source: Own
Let $ a,b>0 $ and $ \frac { a}{b} +\frac {4b}{a}=5. $ Prove that $$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+1}{a}  \geq \frac {7\sqrt 5}{6}  $$
0 replies
1 viewing
sqing
a minute ago
0 replies
D1036 : Composition of polynomials
Dattier   0
5 minutes ago
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
0 replies
Dattier
5 minutes ago
0 replies
inequality
NTssu   4
N 5 minutes ago by Oksutok
Source: Peking University Mathematics Autumn Camp
For given real number $\theta_1, \theta_2, ......, \theta_l$, prove there exists positive integer $k$ and positive real number $a_1, a_2, ......, a_k$, such that $a_1+a_2+ ......+ a_k=1$, for any $n \leq k$, $m \in \{1,2,......,l\}$, $\left| \sum_{j=1}^n a_j sin(j \theta_m ) \right|< \frac{1}{2018n} $ holds.
4 replies
NTssu
Oct 11, 2019
Oksutok
5 minutes ago
Nice geometry
gggzul   0
8 minutes ago
Let $ABC$ be a acute triangle with $\angle BAC=60^{\circ}$. $H, O$ are the orthocenter and excenter. Let $D$ be a point on the same side of $OH$ as $A$, such that $HDO$ is equilateral. Let $P$ be a point on the same side of $BD$ as $A$, such that $BDP$ is equilateral. Let $Q$ be a point on the same side of $CD$ as $A$, such that $CDP$ is equilateral. Let $M$ be the midpoint of $AD$. Prove that $P, M, Q$ are collinear.
0 replies
gggzul
8 minutes ago
0 replies
[Sipnayan JHS] Semifinals Round B, Average, #2
LilKirb   1
N 3 hours ago by LilKirb
How many trailing zeroes are there in the base $4$ representation of $2015!$ ?
1 reply
LilKirb
4 hours ago
LilKirb
3 hours ago
2022 SMT Team Round - Stanford Math Tournament
parmenides51   5
N 4 hours ago by vanstraelen
p1. Square $ABCD$ has side length $2$. Let the midpoint of $BC$ be $E$. What is the area of the overlapping region between the circle centered at $E$ with radius $1$ and the circle centered at $D$ with radius $2$? (You may express your answer using inverse trigonometry functions of noncommon values.)


p2. Find the number of times $f(x) = 2$ occurs when $0 \le x \le 2022 \pi$ for the function $f(x) = 2^x(cos(x) + 1)$.


p3. Stanford is building a new dorm for students, and they are looking to offer $2$ room configurations:
$\bullet$ Configuration $A$: a one-room double, which is a square with side length of $x$,
$\bullet$ Configuration $B$: a two-room double, which is two connected rooms, each of them squares with a side length of $y$.
To make things fair for everyone, Stanford wants a one-room double (rooms of configuration $A$) to be exactly $1$ m$^2$ larger than the total area of a two-room double. Find the number of possible pairs of side lengths $(x, y)$, where $x \in N$, $y \in N$, such that $x - y < 2022$.


p4. The island nation of Ur is comprised of $6$ islands. One day, people decide to create island-states as follows. Each island randomly chooses one of the other five islands and builds a bridge between the two islands (it is possible for two bridges to be built between islands $A$ and $B$ if each island chooses the other). Then, all islands connected by bridges together form an island-state. What is the expected number of island-states Ur is divided into?


p5. Let $a, b,$ and $c$ be the roots of the polynomial $x^3 - 3x^2 - 4x + 5$. Compute $\frac{a^4 + b^4}{a + b}+\frac{b^4 + c^4}{b + c}+\frac{c^4 + a^4}{c + a}$.


p6. Carol writes a program that finds all paths on an 10 by 2 grid from cell (1, 1) to cell (10, 2) subject to the conditions that a path does not visit any cell more than once and at each step the path can go up, down, left, or right from the current cell, excluding moves that would make the path leave the grid. What is the total length of all such paths? (The length of a path is the number of cells it passes through, including the starting and ending cells.)


p7. Consider the sequence of integers an defined by $a_1 = 1$, $a_p = p$ for prime $p$ and $a_{mn} = ma_n + na_m$ for $m, n > 1$. Find the smallest $n$ such that $\frac{a_n^2}{2022}$ is a perfect power of $3$.


p8. Let $\vartriangle ABC$ be a triangle whose $A$-excircle, $B$-excircle, and $C$-excircle have radii $R_A$, $R_B$, and $R_C$, respectively. If $R_AR_BR_C = 384$ and the perimeter of $\vartriangle ABC$ is $32$, what is the area of $\vartriangle ABC$?


p9. Consider the set $S$ of functions $f : \{1, 2, . . . , 16\} \to \{1, 2, . . . , 243\}$ satisfying:
(a) $f(1) = 1$
(b) $f(n^2) = n^2f(n)$,
(c) $n |f(n)$,
(d) $f(lcm(m, n))f(gcd(m, n)) = f(m)f(n)$.
If $|S|$ can be written as $p^{\ell_1}_1 \cdot p^{\ell_2}_2 \cdot ... \cdot  p^{\ell_k}_k$ where $p_i$ are distinct primes, compute $p_1\ell_1+p_2\ell_2+. . .+p_k\ell_k$.


p10. You are given that $\log_{10}2 \approx 0.3010$ and that the first (leftmost) two digits of $2^{1000}$ are 10. Compute the number of integers $n$ with $1000 \le n \le 2000$ such that $2^n$ starts with either the digit $8$ or $9$ (in base $10$).


p11. Let $O$ be the circumcenter of $\vartriangle ABC$. Let $M$ be the midpoint of $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively, onto the opposite sides. $EF$ intersects $BC$ at $P$. The line passing through $O$ and perpendicular to $BC$ intersects the circumcircle of $\vartriangle ABC$ at $L$ (on the major arc $BC$) and $N$, and intersects $BC$ at $M$. Point $Q$ lies on the line $LA$ such that $OQ$ is perpendicular to $AP$. Given that $\angle BAC = 60^o$ and $\angle AMC = 60^o$, compute $OQ/AP$.


p12. Let $T$ be the isosceles triangle with side lengths $5, 5, 6$. Arpit and Katherine simultaneously choose points $A$ and $K$ within this triangle, and compute $d(A, K)$, the squared distance between the two points. Suppose that Arpit chooses a random point $A$ within $T$ . Katherine plays the (possibly randomized) strategy which given Arpit’s strategy minimizes the expected value of $d(A, K)$. Compute this value.


p13. For a regular polygon $S$ with $n$ sides, let $f(S)$ denote the regular polygon with $2n$ sides such that the vertices of $S$ are the midpoints of every other side of $f(S)$. Let $f^{(k)}(S)$ denote the polygon that results after applying f a total of k times. The area of $\lim_{k \to \infty} f^{(k)}(P)$ where $P$ is a pentagon of side length $1$, can be expressed as $\frac{a+b\sqrt{c}}{d}\pi^m$ for some positive integers $a, b, c, d, m$ where $d$ is not divisible by the square of any prime and $d$ does not share any positive divisors with $a$ and $b$. Find $a + b + c + d + m$.


p14. Consider the function $f(m) = \sum_{n=0}^{\infty}\frac{(n - m)^2}{(2n)!}$ . This function can be expressed in the form $f(m) = \frac{a_m}{e} +\frac{b_m}{4}e$ for sequences of integers $\{a_m\}_{m\ge 1}$, $\{b_m\}_{m\ge 1}$. Determine $\lim_{n \to \infty}\frac{2022b_m}{a_m}$.


p15. In $\vartriangle ABC$, let $G$ be the centroid and let the circumcenters of $\vartriangle BCG$, $\vartriangle CAG$, and $\vartriangle ABG$ be $I, J$, and $K$, respectively. The line passing through $I$ and the midpoint of $BC$ intersects $KJ$ at $Y$. If the radius of circle $K$ is $5$, the radius of circle $J$ is $8$, and $AG = 6$, what is the length of $KY$ ?



PS. You should use hide for answers. Collected here.
5 replies
parmenides51
Jun 30, 2022
vanstraelen
4 hours ago
[Sipnayan SHS] Finals Round, Difficult
LilKirb   1
N Today at 6:59 AM by LilKirb
Let $f$ be a polynomial with nonnegative integer coefficients. If $f(1) = 11$ and $f(11) = 2311$, what is the remainder when $f(10)$ is divided by $1000?$
1 reply
LilKirb
Today at 6:49 AM
LilKirb
Today at 6:59 AM
Inequalities
sqing   1
N Today at 3:50 AM by sqing
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
1 reply
sqing
Today at 3:11 AM
sqing
Today at 3:50 AM
Inequalities
sqing   19
N Today at 2:50 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
19 replies
sqing
May 13, 2025
sqing
Today at 2:50 AM
inequality
luckvoltia.112   9
N Today at 2:35 AM by Shan3t
Given that \( a, b, c, d \) are nonzero real numbers, find the minimum value of the expression
\[
P = \left| \frac{b + c + d}{a} \right| + \left| \frac{c + d + a}{b} \right| + \left| \frac{d + a + b}{c} \right| + \left| \frac{a + b + c}{d} \right|.
\]
9 replies
luckvoltia.112
Today at 12:45 AM
Shan3t
Today at 2:35 AM
prove that
luckvoltia.112   0
Today at 12:52 AM
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c > 0 \) and
\[
\frac{25a + 36b + 49c}{5a + 6b + 7c} + \frac{25b + 36c + 49a}{5b + 6c + 7a} + \frac{25c + 36a + 49b}{5c + 6a + 7b} = 18.
\]Prove that exactly two of the numbers \( a, b, c \) are equal to 0.
0 replies
luckvoltia.112
Today at 12:52 AM
0 replies
Geometry Trigonometry Olympiads
Foxellar   0
Yesterday at 11:07 PM
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
0 replies
Foxellar
Yesterday at 11:07 PM
0 replies
Proof of ramsey number
smadadi1000   1
N Yesterday at 10:35 PM by smadadi1000
How do you prove that r(n,2)=n using the pigeonhole principle?
1 reply
smadadi1000
Yesterday at 10:31 PM
smadadi1000
Yesterday at 10:35 PM
Minimize
lgx57   1
N Yesterday at 5:25 PM by Math-lover1
Minimize $\sqrt{\cos^2 x+(2-\sin x)^2}+\dfrac{1}{2}\sqrt{(\sqrt 3-\cos x)^2+(\sin x+1)^2}$
1 reply
lgx57
Yesterday at 1:29 PM
Math-lover1
Yesterday at 5:25 PM
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
Valentin Vornicu   67
N Today at 7:10 AM by alexanderchew
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
67 replies
Valentin Vornicu
Oct 24, 2005
alexanderchew
Today at 7:10 AM
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
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shendrew7
799 posts
#56
Y by
We note $n_1=3^2$ satisfies the condition. We aim to find a prime $p$ such that $pn_i$ satisfies the condition and $\gcd(p,n_i)=1$, so it suffices to have
\[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\]
This primitive root of 2 modulo $p$ exists by Zsigmondy, as desired. Hence we can induct to find $n_{2000}$ with the desired 2000 prime divisors. $\blacksquare$
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peppapig_
280 posts
#57
Y by
Overkill?

In general, I claim that there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$.

I prove this by claiming that if there exists a positive integer $n$ with $k$ distinct integer divisors in the form of
\[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]such that $p_1<p_2\dots<p_k$, then there exists a prime $p_{k+1}>p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want.

C1: First, before we begin, I claim that if any prime $q\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because
\[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]and by LTE (which we can use, since $n\mid 2^n+1$ implies that $q\mid 2^n+1$), this gives us that
\[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]which means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions.

C2: Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that
\[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]and since $ord_r(2)\mid \phi(r)=r-1$, we get that $ord_r(2)\mid \gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with (C1), will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$.

To prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by (C1). Next, by LTE, we have,
\[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]which we can do since we know that $q^c \mid n$ and $r\mid 2^{q^c}+1$, which gives that $r\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\mid 2^{rn}+1$, which combined with (C1), proves our inductive step claim.

Finally, to complete our induction, note that $3\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem.
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SenorSloth
37 posts
#58
Y by
We claim that the answer is yes. We will induct to show that if there exists such an $n$ with $x$ distinct prime factors, then there also exists such an $n$ with $x+1$ distinct prime factors. By repeating this logic we can find an $n$ with exactly $2000$ distinct prime factors.

Our base case is $n=9$, which works since $9\mid 513$. By Zsigmondy, for any $n>3$, there exists a prime $p$ dividing $2^n+1$ that does not divide any $2^k+1$ for smaller $k$. Since this implies that $2$ has order $2n$ modulo $p$, the prime is at least $2n+1$ and thus cannot divide $n$. Thus we know that $pn\mid 2^n+1$ and consequently $pn \mid 2^{pn}+1$. $pn$ has $1$ more distinct prime factor than $n$, so the induction is complete.
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Niku
120 posts
#59
Y by
Do you realise that this post was made 20 years ago.
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BestAOPS
707 posts
#60
Y by
Overkill proof while forgetting about Zsigmondy:

Define two sequences as follows: $n_0 = 1$; $p_i$ is (a) the smallest prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$, or (b) if that doesn't exist, the smallest prime factor of $2^{n_i} + 1$; and $n_{i+1} = n_ip_i.$

Notice that each $n_i$ is divisible by all of the previous ones, and that all $n_i$ and $p_i$ are odd.

First, we show that all $n_i$ satisfy $n_i \mid 2^{n_i} + 1$. We proceed by induction. We can see that $n_0 = 1$ works, so assume $n_i$ works (and all the ones before it). We want to prove $n_{i+1} \mid 2^{n_{i+1}} + 1$.

Note that if a prime $p$ divides $n_{i+1}$, then $p = p_j$ for some $j$ satisfying $0 \leq j \leq i$.
This also means that $p_j$ is a factor of $2^{n_j} + 1$.
Then, by LTE, we have
\[ \nu_{p_j}(2^{n_{i+1}} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}\left(\frac{n_{i+1}}{n_j}\right) = \nu_{p_j}(n_{i + 1}) + \nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j). \]However, the strong inductive hypothesis implies $\nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j) \geq 0$, so we have $\nu_{p_j}(2^{n_{i+1}} + 1) \geq \nu_{p_j}(n_{i+1})$.
As this is true for all $p$, the inductive step is complete.

Next, we claim that eventually, the number of distinct prime factors of $n_i$ is always one more than the number of distinct prime factors of $n_{i-1}$. This is equivalent to showing that eventually, there always exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$.

Suppose, for some $i$, that every prime factor of $2^{n_i} + 1$ is also a factor of $n_i$.
Then, let $p$ be a prime factor of $n_i$, and pick the minimal $j$ such that $p_j = p$. This minimality implies that $\nu_{p_j}(n_j) = 0$.
We have
\[ \nu_{p_j}(2^{n_i} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i) - \nu_{p_j}(n_j) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i). \]We can raise $p_j$ to the power of both sides to get
\[ {p_j}^{\nu_{p_j}(2^{n_i} + 1)} = {p_j}^{\nu_{p_j}(2^{n_j} + 1)} {p_j}^{\nu_{p_j}(n_i)}. \]Doing this for every prime factor $p$ of $n_i$ (notice that $j$ is now a one-to-one function of $p$) and multiplying the resulting equations, we get
\[ 2^{n_i} + 1 = n_i \prod_p p^{\nu_p(2^{n_{j(p)}} + 1)}. \]For all $p$, we have $p^{\nu_p(2^{n_{j(p)}} + 1)} \leq 2^{n_{j(p)}} + 1$. Thus,
\[ 2^{n_i} + 1 \leq n_i \prod_p (2^{n_{j(p)}} + 1). \]Since $j$ is one-to-one, every $j(p)$ is unique and in the set $\{0, 1, \ldots, i-1\}$. Therefore, we have the inequality
\[ 2^{n_i} + 1 \leq n_i \prod_{j=0}^{i-1} (2^{n_j} + 1). \]Taking the log base $2$ of both sides, we have
\[ n_i < \log _2 (2^{n_i} + 1) \leq \log _2 n_i + \sum _{j=0}^{i-1} \log_2(2^{n_j} + 1) < \log_2 n_i + \sum _{j=0}^{i-1} (n_j + 1) = \log_2 n_i + i + \sum_{j=0}^{i-1}n_j. \]Now, in order to achieve a bound on $n_i$, we notice that $p_i \geq 3$ for all $i$, so therefore, $n_i \geq 3^i$. It is then easy to see that $\sum_{j=0}^{i-1}n_j \leq \frac12 n_i$ for all $i \geq 1$. Then,
\[ n_i < \log_2 n_i + i + \frac12 n_i \leq \log_2 n_i + \log_3 n_i + \frac12 n_i \iff n_i < 2(\log _2 n_i + \log _3 n_i). \]This inequality obviously cannot be satisfied as $n_i$ grows large. Thus, eventually, we must have that there exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$. Hence, there must eventually exist an $n$ in our sequence with exactly 2000 distinct prime factors.
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LQFNB
12 posts
#61
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是否存在一个恰有2000个素因子的正整$n$$n \mid 2^n + 1$?
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LQFNB
12 posts
#62
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是否存在一个恰有2000个不同素因子的正整$n$$n \mid 2^n + 1$
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eibc
600 posts
#63
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The answer is yes. We will construct a positive integer $n = p_1^2 p_2p_3 \cdots p_{2000}$ where $p_1 < p_2 < \cdots < p_{2000}$ are distinct primes such that $n \mid 2^n + 1$.

First, we set $p_1 = 3$. For $i \ge 2$, let $q_i = 3^2 \cdot p_2 \cdots p_3 \cdots p_{i - 1}$. Then, we shall construct $p_i$ recursively by picking a primitive divisor of $2^{2q_i} - 1$, which exists by Zsigmondy's theorem.

Now, we show that $n \mid 2^n + 1$. It suffices to show that for $1 \le i \le 2000$, we have $\nu_{p_i}(2^n + 1) \ge \nu_{p_i}(n)$. For $i = 1$, we note that by LTE,
$$\nu_3(2^n + 1) = \nu_3(2 + 1) + \nu_3(n) = 1 + \nu_3(n) > 1.$$For $i > 1$, we note that by construction, $\text{ord}_{p_i}(2) = 2q_i$. Thus, $2^{q_i} \equiv -1 \pmod {p_i}$, so by LTE, we have
$$\nu_{p_i}(2^n + 1) = \nu_{p_i}((2^{q_i})^{n/q_i} + 1) = \nu_{p_i}(2^{q_i} + 1) + \nu_{p_i}(n/q_i) \ge 1 + \nu_{p_i}(n) > \nu_{p_i}(n),$$so we are done.
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Natrium
57 posts
#64
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Lemma. If $a\mid 2^a+1$ and $b\mid 2^b+1$, where $a=3^\alpha a'$, $b=3^\beta b'$, $3 \nmid a', b'$ and $a'$ and $b'$ are coprime, then $ab\mid 2^{ab}+1$.
Proof. Obviously, $a$ and $b$ are odd.
$$2^{ab}\equiv (-1)^b\equiv -1 \pmod{a}$$$$2^{ab}\equiv (-1)^a\equiv -1 \pmod{b}$$By LTE, $v_3(2^{ab}+1)=v_3(ab)+1=\alpha + \beta + 1$. Finally, as $a', b', 3^{\alpha+\beta}\mid 2^{ab}+1$ and all three numbers are pairwise coprime, $ab=a'b'3^{\alpha+\beta}\mid 2^{ab}+1$. $\square$

We construct a sequence of primes $p_2, p_3, \dots, p_{2000}$ in the following manner. Let $p_2=19$ (so that $p_2\mid 2^{3^2}+1$). Inductively assume we have constructed $p_2, p_3, \dots, p_{i}$, for some $i\ge 2$, such that:
$\bullet$ all $p_j$ with $2\le j\le i$ are distinct,
$\bullet$ $p_j>3$ and $p_j\mid2^{3^{i}}+1$ for each $2\le j\le i$.
As $2^{3^{i+1}}+1=(2^{3^i}+1)(4^{3^i}-2^{3^i}+1)$ and the greatest common divisor of the terms in the parenthesis is $3$, we conclude that $4^{3^i}-2^{3^i}+1$ has all its prime divisors distinct from $p_j$ for $2\le j\le i$. As $9\mid 2^{3^i}+1$ and $4^{3^i}-2^{3^i}+1>3$, it has a prime factor $p_{i+1}>3$, so the inductive claim holds for $i+1$ as well.

Having constructed $p_2, p_3, \dots, p_{2000}$, let $n_i=3^i p_i$ for each $2\le i\le 2000$. By LTE, $v_3(2^{n_i}+1)=i+1$. By Fermat's Little Theorem, $2^{n_i}\equiv 2^{3^i p_i}\equiv 2^{3^i}\equiv -1\pmod{p_i}$. Therefore, $n_i = 3^i p_i\mid 2^{n_i} + 1$ for each $i$. Let $n=n_2n_3\dots n_{2000}$. By repeated application of the lemma, $n\mid 2^n + 1$, and by construction, $n$ has $2000$ distinct prime divisors.
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smileapple
1010 posts
#65
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Write $x\sim y$ if $p\mid x$ iff $p\mid y$.

Let $n$ be some odd integer such that $n>3$, $n\mid 2^n+1$, and $n\nsim 2^n+1$. We claim that there exists a prime $p\nmid n$ for which $np\mid 2^{np+1}$ and $np\nsim 2^{np}+1$. Indeed, take some $p\nmid n$ and $p\mid 2^n+1$. Note that $2^{np}+1\equiv 2^n+1\equiv0\pmod p$, and since $p$ is odd, we also have $n\mid 2^{np}+1$. Hence $np\mid 2^{np}+1$. Furthermore, since $n>3$, there exists some prime $q$ satisfying $q\nmid 2^n+1$ and $q\mid 2^{np}+1$ by Zsigmondy.

Applying the above claim $1999$ times on $n=9$ implies that the answer is $\fbox{\text{yes}}$. $\blacksquare$
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amapstob
19 posts
#66
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The answer is yes.
Lemma. For all primes $p>3$, $2^p+1$ has a prime divisor greater than $p$.
Proof. $2^p\equiv -1\pmod{q}\implies 2^{\gcd(q-1,2p)}\equiv 1\pmod{q}$. If $\gcd(q-1,2p)=2$, then $q=3$, so we have to rule out $2^p+1$ being a power of three, which only happens with $p=3$ by Mihailescu's theorem. Since $p>3$, there exists $q\neq 3$ dividing $2^p+1$. Then $p\mid q-1\implies p<q$, as desired. $\blacksquare$

Claim. If $n\mid 2^{n}+1$ and $n$ is odd and has a prime divisor greater than $3$, there exists a prime with $p\mid 2^n+1$ and $p\nmid n$ such that $np\mid 2^{np}+1$.
Proof. Let $q$ be the greatest prime divisor of $n$. Then $2^q+1\mid 2^n+1$. But $2^q+1$ has a prime divisor $p$ greater than $q$ by the above lemma. So $p\nmid n$ and $p\mid 2^n+1$. Then since $p\mid 2^n+1$ and $n\mid 2^n+1$ and $n,p$ are coprime, $np\mid 2^n+1$. But $2^n+1\mid 2^{np}+1$, so we're done. $\blacksquare$

Now observe that $3^2\cdot 19 \mid 2^{3^2\cdot 19}+1$, so applying the lemma above $1998$ times finishes. $\blacksquare$
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cursed_tangent1434
640 posts
#67
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We claim that the answer is yes. In fact we can show the more general statement that for any positive integer $d$ there exists some positive integer $n$ for which $n$ has exactly $d$ distinct prime divisors and $n \mid 2^n+1$. To do this, we employ induction.

First note that $3 \mid 2^3+1$. Now, say there exists a positive integer $n_r = 3^{r}\cdot p_1p_2 \dots p_r$ for which $n_r \mid 2^{n_r}+1$. Consider
\[2^{3n_r}+1=2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1\]Now, by Zsigmondy's Theorem there exists a prime $p_{r+1} \mid 2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1$ but $p_{r+1} \nmid 2^{3^r \dot p_1p_2\dots p_r}$. Thus, this prime factor $p_{r+1} \not \in \{p_1,p_2,\dots , p_r\}$. Further,
\[3n_r \mid 2^{n_r}+1 \mid 2^{3n_r\cdot p_{r+1}}+1\]since by Lifting the Exponent Lemma, $\nu_3(2^{n_r}+1)= \nu_3(3)+\nu_3(n_r) = \nu_3(3n_r)$. Finally,
\[2^{3n_r\cdot p_{r+1}}+1 \equiv 2^{3n_r}+1 \equiv 0 \pmod{p_{r+1}}\]by construction. Thus, $3n_r\cdot p_{r+1} \mid 2^{3n_r\cdot p_{r+1}}+1$ so we can let $n_{r+1}=3n_r\cdot p_{r+1}$ which completes the induction and proves the result.
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ray66
42 posts
#68
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We will prove the result by induction.

First take the base case $n_1=9$ so that $9$ divides $513$. Now consider the number $n_2=n_1p_2$ where $p_2$ is a unique prime number dividing $2^{n_1}+1$. We know that such a $p$ exists by Zsigmondy. Therefore $2^{n_2}+1$ is also divisible by $p_2$, so we finish the induction.
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Ilikeminecraft
658 posts
#69
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I claim that there \boxed{\text{exists}} a number that has $n$ distinct prime numbers that satisfies our conditions.

Let $k_n = 9 \cdot \prod\limits_{i = 1}^{n - 1} p_i$ be a construction for $n$. I will prove this with induction.

Clearly, $k_1 = 9$ is a construction for $n = 1.$

Now, assume that $n = l$ has a valid construction. Let $p_{l}$ be a prime dividing $2^{k_l} + 1$ such that $(p_l, k_l).$ I will prove the existence of such a $p_l:$

Notice that $k$ is not even. We have that $\nu_3(2^{k_l} + 1) = \nu_3(k_l) + \nu_3(3) = 1 + 2 = 3,$ and $$\nu_{p_i}(2^{k_l} + 1) = \nu_{p_i}\left(\left(2^{k_{i + 1}}\right)^{\frac{k_l}{k_{i + 1}}} + 1\right) = \nu_{p_i}(2^{k_{i + 1}} + 1) + \nu_{p_i}\left(\prod_{j = i + 1}^{l - 1}p_j\right) = 1$$Thus, since $3k_{l} < 2^{k_l} + 1,$ there must exist a non-3 value greater than 1 that divides $\frac{2^{k_l} + 1}{k_l}.$ By picking a $p_i$ that divides that, we can gaurantee $p_i$ exists and is relatively prime to all other primes in $k_l.$

Finally, I claim that $k_{l + 1} = k_lp_{l}$ is a valid construction for $n = l + 1.$ We have that $k_l \mid 2^{k_l} + 1 \mid 2^{k_l p_l} + 1,$ and $p_l \mid 2^{k_lp_l} + 1\implies k_{l + 1} = k_lp_l \mid 2^{k_l p_l} + 1 = 2^{k_{l + 1}} + 1.$

Thus, we are done.
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alexanderchew
15 posts
#70
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Evan Chen wrote:
Answer: Yes.

We say that $n$ is Korean if $n \mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim:

Claim: If $n > 3$ is Korean, there exists a prime $p$ not dividing $n$ such that $np$ is Korean too.

Proof. I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, which exists by Zsigmondy theorem. Obviously $p \neq 2$. Then:
  • Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$.
  • Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$;
Hence $np \mid 2^{np} + 1$ by Chinese Theorem, since $\gcd(n,p) = 1$. $\blacksquare$

My solution is almost the same as this but I chose $p|2^n-(-1)^n$ instead
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