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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
Inspired by Kazakhstan 2017
sqing   1
N a few seconds ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a^2+b^2+c^2=2. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge \frac{8}{3}$$$$  \left( 2a^2+\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left( 2a^2+\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)  \ge 9\sqrt[3]{4}$$
1 reply
1 viewing
sqing
a minute ago
sqing
a few seconds ago
Number theory
Ecrin_eren   0
12 minutes ago
Show that there are no prime numbers satisfying the equation

(p + r)^q + (q + r)^p = (p + q)^r.

0 replies
Ecrin_eren
12 minutes ago
0 replies
Wait wasn't it the reciprocal in the paper?
Supercali   7
N 13 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
7 replies
Supercali
Jul 9, 2023
kes0716
13 minutes ago
Functional equation
Ecrin_eren   0
18 minutes ago
Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

0 replies
Ecrin_eren
18 minutes ago
0 replies
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
A great result
steven_zhang123   10
N Yesterday at 9:35 AM by teomihai
Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4} $.
10 replies
steven_zhang123
Oct 31, 2024
teomihai
Yesterday at 9:35 AM
Weird family of sequences
AndreiVila   9
N Yesterday at 8:40 AM by Fibonacci_math
Source: Romanian District Olympiad 2025 12.3
[list=a]
[*] Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
9 replies
AndreiVila
Mar 8, 2025
Fibonacci_math
Yesterday at 8:40 AM
Monster Integral
Entrepreneur   1
N Yesterday at 7:41 AM by RezerdPrime
$$\color{blue}{\int_{0}^{\pi} \frac{\tan^{-1}\left(\frac{\ln(\sin(x))}{x}\right)dx}{\ln^2\left(x^2 + \ln^2(\sin(x))\right) + 4\arctan^2\left(\frac{\ln(\sin(x))}{x}\right)}= -\frac{\pi \tan^{-1}\left(\frac{2\ln(2)}{\pi}\right)}{\ln^2\left(\frac{\pi^2}{4} + \ln^2(2)\right) + 4\arctan^2\left(\frac{2\ln(2)}{\pi}\right)}.}$$
1 reply
Entrepreneur
Jan 10, 2025
RezerdPrime
Yesterday at 7:41 AM
convex closed set with a nonempty interior
ILOVEMYFAMILY   0
Yesterday at 6:11 AM
a) When $n = 2$, prove that a convex closed set with a nonempty interior that contains exactly one extreme point must contain a ray. (SOLVED)

b) When $n = 2$, find all convex closed sets with a nonempty interior that contain exactly one extreme point.
0 replies
ILOVEMYFAMILY
Yesterday at 6:11 AM
0 replies
Differentiation Marathon!
LawofCosine   178
N Yesterday at 5:26 AM by awzhang10
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)
178 replies
LawofCosine
Feb 1, 2025
awzhang10
Yesterday at 5:26 AM
Integrals problems and inequality
tkd23112006   1
N Yesterday at 3:55 AM by removablesingularity
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:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
1 reply
tkd23112006
Feb 16, 2025
removablesingularity
Yesterday at 3:55 AM
International Zhautykov Olympiad 2011 - Problem 5
ybalkas   9
N Jan 21, 2023 by top1vien
Let $n$ be integer, $n>1.$ An element of the set $M=\{ 1,2,3,\ldots,n^2-1\}$ is called good if there exists some element $b$ of $M$ such that $ab-b$ is divisible by $n^2.$ Furthermore, an element $a$ is called very good if $a^2-a$ is divisible by $n^2.$ Let $g$ denote the number of good elements in $M$ and $v$ denote the number of very good elements in $M.$ Prove that
\[v^2+v \leq g \leq n^2-n.\]
9 replies
ybalkas
Jan 17, 2011
top1vien
Jan 21, 2023
International Zhautykov Olympiad 2011 - Problem 5
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ybalkas
31 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $n$ be integer, $n>1.$ An element of the set $M=\{ 1,2,3,\ldots,n^2-1\}$ is called good if there exists some element $b$ of $M$ such that $ab-b$ is divisible by $n^2.$ Furthermore, an element $a$ is called very good if $a^2-a$ is divisible by $n^2.$ Let $g$ denote the number of good elements in $M$ and $v$ denote the number of very good elements in $M.$ Prove that
\[v^2+v \leq g \leq n^2-n.\]
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mavropnevma
15142 posts
#2 • 2 Y
Y by Adventure10, Mango247
1. Let us prove $g \leq n^2 - n$. The number of elements $m \in M$ such that $\gcd(m,n^2) = 1$ is $\varphi(n^2) \geq n$, since for $n = \prod_{k=1}^s p_k^{\alpha_k}$, with $p_k$ primes and $\alpha_k \geq 1$, we have $\varphi(n^2) = \prod_{k=1}^s (p_k -1) \cdot \prod_{k=1}^s p_k^{2\alpha_k - 1} \geq \prod_{k=1}^s p_k^{\alpha_k} = n$. To these elements $m$ thus correspond at least $n-1$ elements $a = m+1 \in M$ (since $n^2-1$ is one of those elements $m$, unavailable for $a$).
As $ab - b = b(a-1)$, when $\gcd(a-1,n^2)=1$ we would need $n^2 \mid b$ in order to have $n^2 \mid ab - b$, impossible since $1\leq b \leq n^2-1$. Therefore the number of good elements is $g \leq (n^2-1) - (n-1) = n^2 -n$ (with equality reached for $n=2$).

Moreover, any time $\gcd(a-1,n^2)=d > 1$, the element $a$ is good, since $b = n^2/d$ is then available, so in conclusion $a$ is good if and only if $\gcd(a-1,n^2) > 1$.
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mavropnevma
15142 posts
#3 • 2 Y
Y by Smoothy, Adventure10
2. Let $a$ be a very good element, so $n^2 \mid a^2 - a = a(a-1)$. Since $\gcd(a,a-1) = 1$, it follows that we must have $a=\alpha x^2$ and $a-1 = \beta y^2$, with $\gcd(x,y) = 1$, $n=xy$ (and also some other restrictions, which are irrelevant). Now assume another very good element $a'$ induces the representation $a'=\alpha' x^2$ and $a'-1 = \beta' y^2$ (using the same $x$ and $y$; if one of them is the same, then the other one needs also be the same).
Then $x^2 \mid a - a'$ and $y^2 \mid (a-1) - (a'-1) = a - a'$, so $n^2 = x^2y^2 \mid |a-a'| < n^2$, therefore $a=a'$, contradiction. As a consequence, the representations of the very good elements $a$ use distinct $x$'s and $y$'s.

Let now $a_1,a_2,\ldots,a_v$ be the $v$ very good elements, with representations $a_k = \alpha_k x_k^2$, indexed such that $1 < x_1 < x_2 < \cdots < x_v$ (we cannot have $x_1 = 1$ since that would imply $y_1 = n$, and so $a_1 - 1 \geq n^2$; and we have seen that the $x_k$ need be distinct). It immediately follows $x_v \geq v+1$.
But then, fixing $b = a_v = \alpha_vx_v^2$, we may choose $a = 1 + \lambda y_v^2$ for all $1 \leq \lambda \leq x_v^2 - 1$, since then $a \leq 1 + (x_v^2 - 1)y_v^2 < n^2$. Moreover, we have $b(a-1) = \alpha_v \lambda x_v^2 y_v^2 = \alpha_v \lambda n^2$, whence $n^2 \mid b(a-1)$. It means all these $a$'s are good, and there are $x_v^2 - 1 \geq (v+1)^2 - 1 = v^2 + 2v$ of them. This means $v^2 + 2v \leq g$, an improvement on the bound asked (and even more maybe could be garnered by considering the other $a_k$ elements, for $1\leq k<v$).
This post has been edited 1 time. Last edited by mavropnevma, Jan 18, 2011, 7:21 AM
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mavropnevma
15142 posts
#4 • 2 Y
Y by Adventure10, Mango247
A small correction (needed, since we may have the case with $v=1$, $g=2$, for example for $n=2$, when $v^2 + v = g$). For an understandable reason (!?!), having reached the crux of the problem, I overlooked that in fact always $x_1 = 1$, since for $a_1 = 1$ we have $n^2 \mid 1(1-1) = 0$. The following fixes this oversight.

In general we have $v\geq 1$ ($1$ is very good) and $g \geq 2$ ($1$ and $n+1$ are good). For $v = 1$ therefore we have $v^2 + v \leq g$ (with equality actually occuring for $n=2$, when $g=2$).
If $v>1$, let us analyze when can we have (in the notations of above) $x_2=2$, i.e. $a_2 = \alpha_2\cdot 4$. This implies $n^2 \mid \alpha_2\beta_2 \cdot 4y_2^2$, thus $n^2 = 4y_2^2$, implying $n=2m$ ($m = y_2$).
Then either $a_2 = m^2 + 1$, with $4 \mid m^2+1$, but then $4 \mid 3m^2 - 1$, so $a = 3m^2 > a_2$ is very good, contradiction, or $a_2 = 3m^2 + 1$, with $4 \mid 3m^2+1$, but then $4 \mid m^2 - 1$, so $1 < a = m^2 < a_2$ is very good, contradiction.
Therefore, when $v>1$, we must have $x_2 > 2$, therefore (by the reasoning in the above) $x_v \geq v+1$, with the above proof leading to the conclusion $v^2 + v < v^2 + 2v \leq g$.
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Tiko93
20 posts
#5 • 2 Y
Y by Adventure10, Mango247
We will prove that g=n^2-f(n^2),where f denotes Euler's totient function.
It is very easy to see that ab-b is div. by n^2( a,b are inM) <=>gcd(a-1,n^2)>1.Such that if gcd(a-1,n^2)=1=> b is div.by n^2,contradiction. And if gcd(a-1,n^2)=d>1,then we can take b=(n^2)/d<=(n^2)/2<=n^2-1 ,such that n>1.Thus b is in M and ab-b is div.by n^2.
Thus we get that g is the number of a's in M s.t. gcd(a-1,n^2)>1.And this number equals to g=(n^2-1)-(f(n^2)-1)=n^2-f(n^2) =n^2-n*f(n)<=n^2-n and equality holds if and only if n=2.
Let find the upper bound of v. if n=(product) p^a .a^2-a is div by n^2 implies that a=(u^2)*s ,and a-1=(v^2*t) ,with gcd (u,v)=1 ,n=uv, s is natural and t is nonnegative.s<v^2,and t<u^2.
Thus u =(prod1)p^a, v=(prod2)p^a ,where the set od prime divisors of u and don't intersect and their union is the set of prime divisors of n.
So we get that (u^2)*s-(v^2)*t=1,it has at most one solution in N,such that if there are 2 such solutions (s,t) and (p,q),then (u^2)*s-(v^2)*t=(u^2)*p-(v^2)*q=>(s-p) is div by v^2. which is impossible ,s.t 1<=s,p<v^2.
Such that for each factoring of n=uv ,we get at most one solution ,then the number of very good elements is at most the number of such factorings.If the number of prime factors of n is k,then we get that this number is 2^k-2.So v<=2^k-2
where v is the number of very good elements.
So v^2+v<=<4^k .
Let prove that g>=4^k).
Such that g=n^2-f(n^2)=n^2-nf(n)>=n^2-n(n-1)=n.
And let prove that n>=4^k.
The last one is true,because we can prove that only finitely many numbers don't satisfy to this inequality.For them we can calculate the value of g .This numbers are 2*3*5 and 2*3*7.

So g>=v^2+v.
This post has been edited 1 time. Last edited by Tiko93, Jan 28, 2011, 11:48 AM
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Tiko93
20 posts
#6 • 2 Y
Y by Adventure10, Mango247
More,we can prove that lim(v/g)=0
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mavropnevma
15142 posts
#7 • 2 Y
Y by Adventure10 and 1 other user
Indeed, $g(n) = n^2 - \varphi(n^2)$, where $\varphi$ is the Euler totient, while $v(n) = 2^{\omega(n)} -1$, where $\omega$ is the arithmetic function counting the distinct primes dividing $n$.
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mmmath
14 posts
#8 • 1 Y
Y by Adventure10
i want to prove that if $n=p_1^{a_1}....P_k^{a_k}$ Then $v=2^k-1$ and $g=n^{2}-\varphi(n^{2})$ then it's easy to prove the inequlity.
$x(x-1)\equiv 0\pmod{p^{2a_i}}$so this equality have to roots so the equality$x(x-1)\equiv 0\pmod{n^2}$have $2^k$roots(by chinese remainder) but $n^2$ is one of the roots so$v=2^k-1$
so we can prove the inequality.
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Assassino9931
1188 posts
#9
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Here is a solution, expanding upon mavropnevma's spoilers.

Regarding $n^2 \mid b(a-1)$, if $\mbox{gcd}((a-1),n^2) = 1$, then $n^2 \mid b$, which is impossible for $1\leq b \leq n^2$; and if $\mbox{gcd}((a-1),n^2) = d>1$, then picking $b=\frac{n^2}{d}$ works. Hence $g$ is equal to $n^2-1$ minus the number of integers in $\{0,1,\ldots,n^2-2\}$ coprime with $n^2$, which is $n^2 - 1 - (\varphi(n^2) - 1) = n^2 - \varphi(n^2)$, where $\varphi$ is the Euler totient. In particular $g\leq n^2 - n$ is equivalent to $\varphi(n^2) \geq n$ which is provable in many ways - either by proving $\varphi(p^{2k}) \geq p^k$ for prime $p$ or just observing that $kn-1$, $k=1,2,\ldots,n$ are coprime with $n$.

Regarding $n^2 \mid a^2 - a$, let $n = \prod_{i=1}^k p_i^{\alpha_i}$ - we equivalently want $p_i^{2\alpha_i} \mid a(a-1)$ for all $i$. As $a$ and $a-1$ are relatively prime, either $a\equiv 0 \pmod {p_i^{2\alpha_i}}$ or $a\equiv 1 \pmod {p_i^{2\alpha_i}}$. So for each $i$ there are two choices for the remainder of $a$ - hence by the Chinese Remainder Theorem there are $2^k$ choices for $a$ modulo $n^2$ overall. As we have to exclude the possibility $a=0$, we deduce $v = 2^k - 1$ and so $v^2 + v = 2^{2k} - 2^k$.

To finish off, we will prove $4^k + \varphi(n^2) \leq n^2$ for $k\geq2$ (for $k=1$ we want $2+\varphi(n^2) \leq n^2$ which is true since $1$ and $n^2-1$ are coprime with $n^2$). Clearly $\varphi(n^2) \leq n^2-n$ (as multiples of $n$ are not coprime with $n$) and so if $n\geq 4^k$ we are done; this certainly holds when $3\nmid n$, as then (recall that $n$ is odd) $n = \prod_{i=1}^k p_i^{\alpha_i} \geq 5^k$. Now suppose $n=3^{\alpha}m$ and $m=\prod_{i=2}^k p_i^{\alpha_i}$. Then we want $4^k + 2\cdot 3^{2\alpha-1}\varphi(m^2) \leq 3^{2\alpha}m^2$ - equivalently, $\frac{4^k}{3^{2\alpha}} + \frac{2}{3}\varphi(m^2) \leq m^2$. With $\varphi(m^2) \leq m^2 - m$ we reduce to $\frac{4^k}{9} \leq \frac{m^2+2m}{3}$, which follows by $m=\prod_{i=2}^k p_i^{\alpha_i} \geq 5^{k-1}$.
This post has been edited 2 times. Last edited by Assassino9931, Jan 3, 2022, 1:50 PM
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top1vien
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#10
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My solution (beautiful problem):
Consider a good number $a$, there exists $b\in M$ which satisfies $n^2\;|\;b(a-1)$
If $(a-1,n^2)=1$ then $n^2\;|\;b$ (contradiction since $1\leq b\leq n^2-1$)
So we must have $(a-1,n^2)>1$
If $(a-1,n^2)=d>1$, choose $b=\frac{n^2}{d}$ then $n^2\;|\;b(a-1)$
So $a$ is good if and only if $(a-1,n^2)>1$
So we get the number of good numbers $a$ which is $n^2-1-(\varphi(n^2)-1)=n^2-\varphi(n^2)$
Notice that: $\varphi(n^2)=\displaystyle n^2\prod_{p\in\wp,p|n}\left(1-\frac{1}{p}\right)=n.n\prod_{p\in\wp,p|n}\left(1-\frac{1}{p}\right)=n\varphi(n)$
So we have $g=n^2-\varphi(n^2)=n^2-n\varphi(n)\leq n^2-n$
Consider the left inequality
Consider a very good number $a$
$p$ is a prime factor of $n$, we have $p^{v_p(n^2)}\;|\;a(a-1)$
But $(a,a-1)=1$ so $p^{v_p(n^2)}\;|\;a$ or $p^{v_p(n^2)}\;|\;a-1$
If $p^{v_p(n^2)}\;|\;a$ for all $p|n$ then $n^2\;|\;a$ (contradiction since $1\leq a\leq n^2-1$)
So we must have $a$ is very good if and only if for all prime factor $p$ of $n$ then $p^{v_p(n^2)}\;|\;a$ or $a-1$, also there must exists a prime factor $p$ of $n$ such that $p^{v_p(n^2)}\;|\;a-1$
Call $r$ the number of distinct prime factor of $n$
We can evaluate $v=2^r-1$ (simple counting)
The problem becomes: Prove that $2^r(2^r-1)\leq n^2-\varphi(n^2)$
Induction on $r$. For $r=1$ the problem is trivial
If the problem is true for some $r$, then $2^r(2^r-1)\leq n^2-\varphi(n^2)$, for all positive integer $n$ which have $r$ distinct prime factors
Consider a prime number $q$ with $(q,n)=1$
We need to prove that $2^{r+1}(2^{r+1}-1)\leq q^{2\alpha}n^2-\varphi(q^{2\alpha}n^2)$ with some $\alpha\in\mathbb Z^+$, but it is true since
$
        q^{2\alpha}n^2\left(1-\prod_{p\in\wp,p|qn}\left(1-\frac{1}{p}\right)\right)\geq q^{2}n^2\left(1-\prod_{p\in\wp,p|qn}\left(1-\frac{1}{p}\right)\right)$
$=q^2n^2-\varphi(q^2n^2)$
$=q^2(n^2-\varphi(n^2))+q^2\varphi(n^2)-q(q-1)\varphi(n^2)$
$\geq q^2.2^r(2^r-1)+q\varphi(n^2)$
$\geq 4.2^r(2^r-1)+2n\varphi(n)$
$\geq 4.2^r(2^r-1)+2^{r+1}$ (since $n$ has $r$ distinct prime factors)
$=2^{r+1}(2^{r+1}-1)$
Combining the 2 inequalities we have $m^2+m\leq k\leq n^2-n$, as desired
The equality happens, for example $n=2$
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