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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
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 4 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
4 minutes ago
H not needed
dchenmathcounts   44
N 38 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
38 minutes ago
IZHO 2017 Functional equations
user01   51
N an hour ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
an hour ago
chat gpt
fuv870   2
N an hour ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
an hour ago
Putnam 2017 B3
goveganddomath   34
N Today at 4:16 AM by OronSH
Source: Putnam
Suppose that $$f(x) = \sum_{i=0}^\infty c_ix^i$$is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.
34 replies
goveganddomath
Dec 3, 2017
OronSH
Today at 4:16 AM
3-dimensional matrix system
loup blanc   1
N Today at 3:33 AM by alexheinis
Let $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$.
i) Find the matrices $B\in M_3(\mathbb{R})$ s.t. $A^TA=B^TB,AA^T=BB^T$.
EDIT. ii) Show that each solution of i) is in $M_3(K)$, where $K=\mathbb{Q}[\cos(\dfrac{2}{3}\arctan(3\sqrt{3}))]$.
iii) Solve i) when $B\in M_3(\mathbb{C})$.
EDIT. In iii) we again consider the transpose of $B$ and not its conjugate transpose.
1 reply
loup blanc
Yesterday at 1:04 PM
alexheinis
Today at 3:33 AM
Square of a rational matrix of dimension 2
loup blanc   9
N Today at 1:28 AM by ysharifi
The following exercise was posted -two months ago- on the Website StackExchange; cf.
https://math.stackexchange.com/questions/5006488/image-of-the-squaring-function-on-mathcalm-2-mathbbq
There was no solution on Stack.

-Statement of the exercise-
We consider the matrix function $f:X\in M_2(\mathbb{Q})\mapsto X^2\in M_2(\mathbb{Q})$.
Find the image of $f$.
In other words, give a method to decide whether a given matrix has or does not have at least a square root
in $M_2(\mathbb{Q})$; if the answer is yes, then give a method to calculate at least one of its roots.
9 replies
loup blanc
Feb 17, 2025
ysharifi
Today at 1:28 AM
Differentiation Marathon!
LawofCosine   181
N Today at 12:24 AM by Levieee
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
181 replies
LawofCosine
Feb 1, 2025
Levieee
Today at 12:24 AM
Integrals problems and inequality
tkd23112006   2
N Yesterday at 6:02 PM by PolyaPal
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
2 replies
tkd23112006
Feb 16, 2025
PolyaPal
Yesterday at 6:02 PM
real analysis
ay19bme   1
N Yesterday at 5:30 PM by alexheinis
.........
1 reply
ay19bme
Yesterday at 3:05 PM
alexheinis
Yesterday at 5:30 PM
Proving an inequality involving cosine functions
pii-oner   3
N Yesterday at 3:33 PM by pii-oner
Hello AoPS Community,

I am curious about how to demonstrate the following inequality:

[code] \sqrt{1 - |\cos(x \pm y)|^a} \leq \sqrt{1 - |\cos(x)|^a} + \sqrt{1 - |\cos(y)|^a}, \quad \text{for } a \geq 1. [/code]

I’ve plotted the functions, and the inequality seems to hold. However, I am looking for a rigorous way to prove it.

I’d greatly appreciate insights into how to break this down analytically and references to similar problems or techniques that might be helpful.

Thank you so much for your guidance and support!
3 replies
pii-oner
Jan 22, 2025
pii-oner
Yesterday at 3:33 PM
Linear algebra
Dynic   2
N Yesterday at 3:32 PM by loup blanc
Let A and B be two square matrices with the same size. Prove that if AB is an invertible matrix, then A and B are also invertible matrices
2 replies
Dynic
Yesterday at 2:48 PM
loup blanc
Yesterday at 3:32 PM
Very hard group theory problem
mathscrazy   3
N Yesterday at 9:50 AM by quasar_lord
Source: STEMS 2025 Category C6
Let $G$ be a finite abelian group. There is a magic box $T$. At any point, an element of $G$ may be added to the box and all elements belonging to the subgroup (of $G$) generated by the elements currently inside $T$ are moved from outside $T$ to inside (unless they are already inside). Initially $
T$ contains only the group identity, $1_G$. Alice and Bob take turns moving an element from outside $T$ to inside it. Alice moves first. Whoever cannot make a move loses. Find all $G$ for which Bob has a winning strategy.
3 replies
mathscrazy
Dec 29, 2024
quasar_lord
Yesterday at 9:50 AM
limit of u(pi/45)
EthanWYX2009   0
Yesterday at 7:08 AM
Source: 2025 Pi Day Challenge T5
Let \(\omega\) be a positive real number. Divide the positive real axis into intervals \([0, \omega)\), \([\omega, 2\omega)\), \([2\omega, 3\omega)\), \([3\omega, 4\omega)\), \(\ldots\), and color them alternately black and white. Consider the function \(u(x)\) satisfying the following differential equations:
\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]with the initial conditions:
\[
u(0) = 1, \quad u'(0) = 1,
\]and the continuity conditions:
\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]Show that
\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]
0 replies
EthanWYX2009
Yesterday at 7:08 AM
0 replies
number theory from China TST
HuangZhen   17
N Jan 25, 2022 by PSS_73939133
Source: China TST 4 Problem 4
Given integer $d>1,m$,prove that there exists integer $k>l>0$, such that $$(2^{2^k}+d,2^{2^l}+d)>m.$$
17 replies
HuangZhen
Mar 22, 2017
PSS_73939133
Jan 25, 2022
number theory from China TST
G H J
G H BBookmark kLocked kLocked NReply
Source: China TST 4 Problem 4
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HuangZhen
29 posts
#1 • 6 Y
Y by rightways, laegolas, anantmudgal09, Hypernova, Adventure10, Mango247
Given integer $d>1,m$,prove that there exists integer $k>l>0$, such that $$(2^{2^k}+d,2^{2^l}+d)>m.$$
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guangzhou-2015
776 posts
#2 • 1 Y
Y by Adventure10
fantastic problem
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Erkhes
143 posts
#3 • 3 Y
Y by Hypernova, Adventure10, Mango247
solution pls
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ABCDE
1963 posts
#4 • 2 Y
Y by Adventure10, Mango247
Take $l>\log_2d$ and consider a prime factor $p$ of $2^{2^l}+d$ such that $p\not\equiv1\pmod{2^l}$. Such a $p$ must exist, or $2^{2^l}+d\equiv1\pmod{2^l}\implies 2^l\mid d-1$, a contradiction as $1<d<2^l$. Now, let $p-1=2^ab$ for some odd $b$ and $a<l$ and set $k=l+\text{ord}_b2$. We now have that $p-1\mid 2^k-2^l$ as it is divisible by both $2^a$ and $b$, so $p\mid 2^{2^k}-2^{2^l}$ and $p\mid 2^{2^k}+d$. Now it suffices to show that there are arbitrarily large primes that divide a number of the form $2^{2^n}+d$, as by the above argument we may take such a large prime greater than $m$ that divides $2^{2^N}+d$ for some $N>l$. This is a well-known Iran TST problem, and you can see here: https://artofproblemsolving.com/community/c6h275813p1492629 for the proof.
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Lam.DL.01
113 posts
#5 • 2 Y
Y by Adventure10, Mango247
You can prove this method: Prove that 2^(2^l)+d have infinity primes as its divisors when l goes to infinity
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tenplusten
1000 posts
#6 • 2 Y
Y by Adventure10, Mango247
Just Kobayashi's theorem.
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MathPanda1
1135 posts
#8 • 1 Y
Y by Adventure10
Unless I am mistaken, it seems that all of the proofs above only prove that there are infinitely many primes dividing the terms. However, there can be infinitely many primes dividing the terms, just each prime divides at most one term. So, it is possible that there are only finitely many primes $\not\equiv 1 \pmod { 2^l}$, in which case ABCDE's solution falls apart at the end? I am sorry if I misunderstood your proofs, and please correct me if I am wrong, since I could this infinite prime case the hardest part of this problem.
This post has been edited 1 time. Last edited by MathPanda1, Mar 23, 2017, 4:57 AM
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andria
824 posts
#9 • 2 Y
Y by Adventure10, Mango247
ABCDE wrote:
$2^{2^l}+d\equiv1\pmod{2^l}\implies 2^l\mid d-1$

Obviously this part is wrong if $2\mid d$. :)
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siddigss
224 posts
#10 • 2 Y
Y by Kaguyaximei, Adventure10
Let $S=\{ p \textrm { a prime } : p|2^{2^n}+d \textrm { for some n } \}$. $S$ is infinite. Assume that there is a positive integer $m$ such that $(2^{2^k}+d,2^{2^l}+d)< m$ for all pairs $(k,l)$ of distinct positive integers. Let $S_1=\{p\in S|p< m\}$ and $S_2=\{p\in S|p\geq m\}$. There is a positive integer $X$ such that $v_p(2^{2^n}+d)<X$ for all $n\in\mathbb{N}$ and $p\in S_1$, otherwise there is a prime in $S_1$ with a very large power that divides two distinct terms of $2^{2^n}+d$. Any prime divisor of $S_2$ divide $2^{2^n}+d$ for exactly one $n$. Let $p\in S_2$ a divisor of $2^{2^n}+d$. Note that the congruence $2^k\equiv 2^n\pmod{p-1}$ has infinitely many solution $k>n$ if $2^{n+1}\not|p-1$. Therefore by our assumption we must have $p\equiv 1\pmod{2^{n+1}}$. Take $n$ very large so that $2^{n+1}>d+\left(\prod_{q\in S_1}q\right)^X$. Let $\alpha_q$ be the largest power of $q$ that divide $2^{2^n}+d$. By our previous observation $d\equiv \prod_{q\in S_1} q^{\alpha_q}\pmod {2^{n+1}}$ from which we find $d=\prod_{q\in S_1}q^{\alpha_q}$, since $d>1$, there is a $q\in S_1$ such that $\alpha_q>0$. but then $q|2$. Therefore $d=2^t$, where $t\in\mathbb{N}$.
Write $t=2^uv$ for some odd $v$. It is not hard to prove that the sequence $2^{l-u}-v$ has infinitely many primes dividing at least one of its terms. Let $r>m$ be a prime that divides a term, say $2^{l-u}-v$, with large $l$. Let $k=l+r-1$. Then $2^ur|(2^l-t,2^k-t)$. Therefore $2^{2^ur}+1|(2^{2^l-t}+1,2^{2^k-t}+1)|(2^{2^m}+2^t,2^{2^k}+2^t)$, but clearly $2^{2^ur}+1>r>m$. Contradiction !
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andria
824 posts
#11 • 9 Y
Y by rightways, WizardMath, tenplusten, rkm0959, Kaguyaximei, Aniruddha07, sabkx, Adventure10, Mango247
Lemma 1: the set of prime divisors of $n^k+c$ is infinite.

proof

Lemma 2: if $a>1$ then $\text{gcd}(a^m+1,a^n+1)=a^{\text{gcd}(m,n)}+1$ for odd $m,n$.

Consider a large enough integer $l$ then if we can find a prime $p$ such that:
$$v_2(p-1)\leq l\ ,\ p>m\ ,\ p\mid 2^{2^l}+d\ \clubsuit$$then:
$$2^{l}\stackrel{p-1}{\equiv}2^{l+\phi(p-1)}\Longrightarrow 2^{2^{l}}\stackrel{p}{\equiv}2^{2^{l+\phi(p-1)}}\Longrightarrow 2^{2^{l}}+d\stackrel{p}{\equiv}2^{2^{l+\phi(p-1)}}+d\Longrightarrow \text{gcd}(2^{2^{l}}+d,2^{2^{l+\phi(p-1)}}+d)\geq p>m$$So assume that there isn't any prime with these conditions ($\clubsuit$). hence for any sufficiently large integers $l$ the prime factorization of $2^{2^l}+d$ is in the following form:
$$2^{2^{l}}+d=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}(2^{l+1}s_1+1)^{\beta_1}(2^{l+1}s_2+1)^{\beta_2} \dots (2^{l+1}s_t+1)^{\beta_t}$$where $p_1,p_2,\dots ,p_k<m$.
Assume $M=p_1^{a_1+1}p_2^{a_2+1}\dots p_k^{a_k+1}$. Since $l$ is sufficiently large for any $s$:
$$2^{l+\phi(\phi(M))s}\stackrel{\phi(M)}{\equiv}2^{l}\Longrightarrow 2^{2^{l+\phi(\phi(M))s}}\stackrel{M}{\equiv}2^{2^{l}}\Longrightarrow 2^{2^{l+\phi(\phi(M))s}}+d\stackrel{M}{\equiv}2^{2^{l}}+d\stackrel{M}{\equiv} p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}(2^{l+1}s_1+1)^{\beta_1}(2^{l+1}s_2+1)^{\beta_2} \dots (2^{l+1}s_t+1)^{\beta_t}$$Hence for any $1\leq i\leq k$ we have:
$$v_{p_i}(2^{2^{l+\phi(\phi(M))s}}+d)=a_i$$So we conclude that:
$$2^{2^{l+\phi(\phi(M))s}}+d=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}(2^{l+\phi(\phi(M))s+1}u_1+1)^{c_1}(2^{l+\phi(\phi(M))s+1}u_2+1)^{c_2} \dots (2^{l+\phi(\phi(M))s+1}u_r+1)^{c_r}$$now observe that:
$$d\stackrel{2^{l+\phi(\phi(M))s+1}}{\equiv}2^{2^{l+\phi(\phi(M))s}}+d\stackrel{2^{l+\phi(\phi(M))s+1}}{\equiv}p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}\Longrightarrow 2^{l+\phi(\phi(M))s+1}\mid d-p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$$Hence:
$$d=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$$then for any $i$ such that $a_i\geq 1$ we have:
$\left.\begin{array}{ccc}
p_i\mid d\\ \\
p_i\mid 2^{2^l}+d=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}(2^{l+1}s_1+1)^{\beta_1}(2^{l+1}s_2+1)^{\beta_2} \dots (2^{l+1}s_t+1)^{\beta_t}
\end{array}\right\}\Longrightarrow p_i\mid 2^{2^{l}}\Longrightarrow p_i=2$

Hence $d=2^s$ ($t\geq 1$).

hence for any $m$ we have:

$$2^{2^m-s}+1=(2^{m+1}s_1+1)^{j_1}\dots (2^{m+1}s_u+1)^{j_w}$$
take a prime $p\mid 2^{2^m-s}+1$. observe that from lemma there is a large number $q$ and $m$ such that:

$\left.\begin{array}{ccc}
q\mid  2^m-s\\ \\
q\mid 2^{m+(q-1)h}-s
\end{array}\right\}\stackrel{(2)}{\Longrightarrow} 2^{2^{v_2(s)}q}+1\mid 2^{2^m-s}+1,2^{2^{v_2(s)}q}+1\mid 2^{2^{m+(q-1)h}-s}+1\Longrightarrow \text{gcd}(2^{2^m}+2^s,2^{2^{m+(q-1)h}}+2^s)\geq 2^{2^{v_2(s)}q}+1>m$ (since $q$ is large enough).

Q.E.D
This post has been edited 2 times. Last edited by andria, Mar 23, 2017, 1:16 PM
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WizardMath
2487 posts
#12 • 3 Y
Y by Gluncho, Adventure10, Mango247
Just to note that Lemma 1 above is a straight consequence of Kobayashi's theorem.
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hutu683
112 posts
#13 • 1 Y
Y by Adventure10
WizardMath wrote:
Just to note that Lemma 1 above is a straight consequence of Kobayashi's theorem.

What's Kobayashi's theorem, please? Can't find it on the Internet.
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babu2001
402 posts
#14 • 1 Y
Y by Adventure10
$\text{Kobayashi's Theorem}$

Consider a sequence $\{a_n\}_{n\in\mathbb{N}}$ which has only finitely many primes dividing atleast one of its terms. Then the sequence $\{a_n+c\}_{n\in\mathbb{N}}\forall c\in\mathbb{Z}$ has infinitely many primes dividing atleast one of its terms.
This post has been edited 2 times. Last edited by babu2001, Apr 14, 2017, 5:15 PM
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hutu683
112 posts
#15 • 2 Y
Y by Adventure10, Mango247
babu2001 wrote:
$\text{Kobayashi's Theorem}$

Consider a sequence $\{a_n\}_{n\in\mathbb{N}}$ which has only finitely many primes dividing atleast one of its terms. Then the sequence $\{a_n+c\}_{n\in\mathbb{N}}\forall c\in\mathbb{Z}$ has infinitely many primes dividing atleast one of its terms.

Thank you. But $c$ can't be zero. :)
This post has been edited 1 time. Last edited by hutu683, Apr 14, 2017, 2:49 PM
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ACGNmath
327 posts
#18 • 3 Y
Y by mijail, JG666, Ru83n05
Let $a_n=2^{2^n}+d$. Suppose there exist $m$ and $d$ such that $(a_k,a_l)\leq m$ for all $k$ and $l$. It follows that if $p^t$ divides $a_k$ and $a_l$ for some positive integers $t$, $k$ and $l$, then $p^t<m$. In other words, $\nu_p(a_k)$ is bounded.

Lemma 1: If positive integers $n$ and $k$ satisfy $\nu_2(k)<n$, then $k|2^l-2^n$ for some $l>n$.
Proof: Let $k=2^a b$, where $b$ is odd. We have $a<n$. There exists $m$ such that $b|2^m-1$. Then $$2^ab|2^{n+m}-2^n=2^n(2^m-1)$$
Lemma 2: If $p|a_n$ for some prime $p>m$ and integral $n$, then $p\equiv 1\pmod{2^n}$.
Proof: Suppose $2^n$ does not divide $p-1$. Then it follows from lemma 1 that $p-1$ divides $2^l-2^n$ for some $l>n$. Then
$$a_l-a_n=2^{2^l}-2^{2^n}=2^{2^n}(2^{2^l-2^n}-1)$$The quantity $2^{2^l-2^n}-1$ is divisible by $p$ because $2^l-2^n$ is divisible by $p-1$. Then $p\leq (a_n,a_l)\leq m$, contradiction.

Lemma 3: $d$ is a power of $2$
Proof: For each $n$ we write $a_n=2^{k_n}b_nc_n$, where $b_n$ contains only odd prime divisors of $a_n$ that are less than $m$ and $c_n$ contains those that are not less than $m$. It follows from Lemma 2 that $c_n\equiv 1\pmod{2^n}$, therefore $a_n\equiv d\equiv 2^{k_n}b_n\pmod{2^n}$. The number of primes less than $m$ is finite and for each of them the sequence $\nu_p(a_n)$ is bounded. Moreover, $k_n=\nu_2(d)$ when $n>\nu_2(d)$. This means there exists $M$ such that $2^{k_n}b_n<M$ for every integer $n$, that is, $2^{k_n}b_n=d$ for sufficiently large $n$.
Thus $d$ divides $a_n=2^{2^n}+d$ for sufficiently large $n$, which means $d$ is a power of $2$.

Lemma 4: For large enough $n$, there exists $l>n$ such that $a_n|a_l$.
Proof: Let $d=2^k$. Choose $n>\nu_2(k)$. Then $\nu_2(2^n-k)=\nu_2(k)$. It follows from Lemma 1 that there exists $l$ such that $2^n-k|2^l-2^n$ and therefore $2^n-k|2^l-k$.
Since $\nu_2(2^l-k)=\nu_2(k)=\nu_2(2^n-k)$, the number $\frac{2^l-k}{2^n-k}$ is odd, and thus $2^{2^n-k}+1$ divides $2^{2^l-k}+1$. Multiplying by $2^k$, we get that $a_n|a_l$.

It follows from Lemma 4 that $a_n=(a_n,a_l)\leq m$ for each $n>\nu_2(k)$, a contradiction.
This post has been edited 1 time. Last edited by ACGNmath, Jun 25, 2020, 6:42 AM
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Gaussian_cyber
162 posts
#20 • 5 Y
Y by Juanscholtze, Nathanisme, k12byda5h, JG666, sabkx
Soln
This post has been edited 2 times. Last edited by Gaussian_cyber, Sep 11, 2020, 7:31 PM
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SerdarBozdag
892 posts
#21
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ACGNmath wrote:
Lemma 2: If $p|a_n$ for some prime $p>m$ and integral $n$, then $p\equiv 1\pmod{2^n}$.
Proof: Suppose $2^n$ does not divide $p-1$. Then it follows from lemma 1 that $p-1$ divides $2^l-2^n$ for some $l>n$. Then
$$a_l-a_n=2^{2^l}-2^{2^n}=2^{2^n}(2^{2^l-2^n}-1)$$The quantity $2^{2^l-2^n}-1$ is divisible by $p$ because $2^l-2^n$ is divisible by $p-1$. Then $p\leq (a_n,a_l)\leq m$, contradiction.

$p|2^{2^l-2^n}-1$ does not imply $p|2^{2^n}+d, 2^{2^l}+d$. How did you get contradiction?
This post has been edited 1 time. Last edited by SerdarBozdag, Sep 4, 2021, 7:22 PM
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PSS_73939133
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#22
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(geometric)
This post has been edited 1 time. Last edited by PSS_73939133, Jan 27, 2022, 10:03 AM
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