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IMO ShortList 2001, number theory problem 4
orl   43
N an hour ago by Zany9998
Source: IMO ShortList 2001, number theory problem 4
Let $p \geq 5$ be a prime number. Prove that there exists an integer $a$ with $1 \leq a \leq p-2$ such that neither $a^{p-1}-1$ nor $(a+1)^{p-1}-1$ is divisible by $p^2$.
43 replies
orl
Sep 30, 2004
Zany9998
an hour ago
Number theory - Iran
soroush.MG   32
N an hour ago by Nobitasolvesproblems1979
Source: Iran MO 2017 - 2nd Round - P1
a) Prove that there doesn't exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: gcd(a_i+j,a_j+i)=1$

b) Let $p$ be an odd prime number. Prove that there exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: p \not | gcd(a_i+j,a_j+i)$
32 replies
soroush.MG
Apr 20, 2017
Nobitasolvesproblems1979
an hour ago
Inspired by my own results
sqing   2
N 2 hours ago by cazanova19921
Source: Own
Let $ a,b,c\geq \frac{1}{2}  . $ Prove that
$$ (a+1)(b+2)(c +1)-15 abc\leq \frac{15}{4}$$$$ (a+1)(b+3)(c +1)-21abc\leq \frac{21}{4}$$$$(a+2)(b+1)(c +2)-25a b c \leq \frac{25}{4}$$$$ (a+2)(b+3)(c +2)-35a b c \leq  \frac{35}{2}$$$$    (a+3)(b+1)(c +3)-49a b c \leq  \frac{49}{4}$$$$ (a+3)(b+2)(c +3)-49a b c \leq \frac{49}{2}$$
2 replies
sqing
3 hours ago
cazanova19921
2 hours ago
Line through incenter tangent to a circle
Kayak   32
N 2 hours ago by L13832
Source: Indian TST D1 P1
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
32 replies
Kayak
Jul 17, 2019
L13832
2 hours ago
D1015 : A strange EF for polynomials
Dattier   3
N 2 hours ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
3 replies
Dattier
Mar 16, 2025
Dattier
2 hours ago
Inequalities
sqing   25
N 2 hours ago by DAVROS
Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=1. $ Prove that
$$(a^2-a+1)(b^2-b+1) \geq 9$$$$ (a^2-a+b+1)(b^2-b+a+1) \geq 25$$Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=\frac{2}{3}. $ Prove that
$$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$$$$(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$$
25 replies
sqing
Mar 10, 2025
DAVROS
2 hours ago
Turkey EGMO TST 2017 P6
nimueh   4
N 2 hours ago by Nobitasolvesproblems1979
Source: Turkey EGMO TST 2017 P6
Find all pairs of prime numbers $(p,q)$, such that $\frac{(2p^2-1)^q+1}{p+q}$ and $\frac{(2q^2-1)^p+1}{p+q}$ are both integers.
4 replies
nimueh
Jun 1, 2017
Nobitasolvesproblems1979
2 hours ago
An inequality
JK1603JK   4
N 2 hours ago by Quantum-Phantom
Source: unknown
Let a,b,c>=0: ab+bc+ca=3 then maximize P=\frac{a^2b+b^2c+c^2a+9}{a+b+c}+\frac{abc}{2}.
4 replies
JK1603JK
Yesterday at 10:28 AM
Quantum-Phantom
2 hours ago
Inspired by Abelkonkurransen 2025
sqing   1
N 2 hours ago by kiyoras_2001
Source: Own
Let $ a,b,c $ be real numbers such that $  a^2+4b^2+16c^2= abc. $ Prove that $$\frac{1}{a}+\frac{1}{2b}+\frac{1}{4c}\geq -\frac{1}{16}$$Let $ a,b,c $ be real numbers such that $ 4a^2+9b^2+16c^2= abc. $ Prove that $$ \frac{1}{2a}+\frac{1}{3b}+\frac{1}{4c}\geq -\frac{1}{48}$$
1 reply
sqing
Yesterday at 1:06 PM
kiyoras_2001
2 hours ago
Inequalities
sqing   11
N 2 hours ago by DAVROS
Let $ a,b $ be real numbers such that $ a + b  \geq  |ab + 1|. $ Prove that$$ a^3 + b^3 \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 2(a + b ) \geq  |ab + 1|. $ Prove that$$26( a^3 + b^3) \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$
11 replies
sqing
Mar 8, 2025
DAVROS
2 hours ago
Geometry challenging question
srnjbr   0
3 hours ago
Given a triangle ABC. A1, B1 and C1 are the points of contact of the inner circumcircle of the triangle with the sides BC, AC and AB respectively. The point of contact of AA1 with B1C1 and the circumcircle are called L and Q respectively. M is the midpoint of B1C1. The point of intersection of lines BC and B1C1 is called T. P is the foot of the perpendicular drawn to AT from point L. Show that points A1, M, Q and P lie on a circle.
0 replies
srnjbr
3 hours ago
0 replies
Plane normal to vector
RenheMiResembleRice   0
3 hours ago
Source: Bian Wei
Solve the attached
0 replies
RenheMiResembleRice
3 hours ago
0 replies
Inequality with integers and indices
Michael Niland   7
N Mar 18, 2025 by Ninjametry
Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$, and for a positive integer $n\geq3$, prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$.
7 replies
Michael Niland
Mar 17, 2025
Ninjametry
Mar 18, 2025
Inequality with integers and indices
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Michael Niland
681 posts
#1
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Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$, and for a positive integer $n\geq3$, prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$.
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invisibleman
13 posts
#2
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In order to solve the first problem, I have the following idea: raise both sides to the 6th power, and you're done! Right?
In the general case, consider the function f(x)=lnx/x, whose derivative is negative for x>e, so the function is decreasing. Let x=n and x=n+1, and compare them! I think you understand, if not, write and I'll answer!
This post has been edited 3 times. Last edited by invisibleman, Mar 18, 2025, 11:36 AM
Reason: typo
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krish6_9
18 posts
#3
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The inequality is equivalent to $n^{n+1} > (n+1)^n$. Taking natural logarithm on both sides, this is equivalent to $(n+1) \ln n > n \ln (n+1).$ Dividing, this is equivalent to $\frac{n+1}{\ln (n+1)} > \frac{n}{\ln n}$. Since $\frac{x}{\ln x}$ is increasing (obviously linear functions increase more quickly than logarithms, or using the derivative like @above said), we are done by definition.

i hope this is right! :D :ddr:
well someone taught me the solution hehe
This post has been edited 2 times. Last edited by krish6_9, Mar 18, 2025, 12:31 AM
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Speedysolver1
80 posts
#4
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Help me with a forum
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krish6_9
18 posts
#5
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who, me? wut does that mean
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ohiorizzler1434
718 posts
#6
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Bro! Let's show n^(n+1) > (n+1)^n , or (1+1/n)^n < n. Note that (1+1/n)^n < e because this limits upwards towards e. So for n >= 3, this is obviously true as e<3. Bruh! Now that's rizz!
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allisok100
1 post
#7
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f(x)=\frac{\ln x}{x}\Rightarrow f'(x)=\frac{1-\ln x}{{{x}^{2}}}<0,\forall x>e\Rightarrow f(n)>f(n+1),\forall n\ge 2
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Ninjametry
284 posts
#8
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Graph x^(1/x)
now observe where we get maxima
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