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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
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   1
N a few seconds ago by Ritwin
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
1 reply
+1 w
Primeniyazidayi
12 minutes ago
Ritwin
a few seconds ago
Easy Taiwanese Geometry
USJL   14
N 23 minutes ago by Want-to-study-in-NTU-MATH
Source: 2024 Taiwan Mathematics Olympiad
Suppose $O$ is the circumcenter of $\Delta ABC$, and $E, F$ are points on segments $CA$ and $AB$ respectively with $E, F \neq A$. Let $P$ be a point such that $PB = PF$ and $PC = PE$.
Let $OP$ intersect $CA$ and $AB$ at points $Q$ and $R$ respectively. Let the line passing through $P$ and perpendicular to $EF$ intersect $CA$ and $AB$ at points $S$ and $T$ respectively. Prove that points $Q, R, S$, and $T$ are concyclic.

Proposed by Li4 and usjl
14 replies
+1 w
USJL
Jan 31, 2024
Want-to-study-in-NTU-MATH
23 minutes ago
Problem 7
SlovEcience   6
N 32 minutes ago by Li0nking
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
6 replies
SlovEcience
May 14, 2025
Li0nking
32 minutes ago
Strange circles in an orthocenter config
VideoCake   1
N 32 minutes ago by KrazyNumberMan
Source: 2025 German MO, Round 4, Grade 12, P3
Let \(\overline{AD}\) and \(\overline{BE}\) be altitudes in an acute triangle \(ABC\) which meet at \(H\). Suppose that \(DE\) meets the circumcircle of \(ABC\) at \(P\) and \(Q\) such that \(P\) lies on the shorter arc of \(BC\) and \(Q\) lies on the shorter arc of \(CA\). Let \(AQ\) and \(BE\) meet at \(S\). Show that the circumcircles of \(BPE\) and \(QHS\) and the line \(PH\) concur.
1 reply
VideoCake
Monday at 5:10 PM
KrazyNumberMan
32 minutes ago
Largest Prime Factor
P162008   3
N 3 hours ago by maromex
The largest prime factor of the sum $\sum_{k=1}^{11} k^5$ is $\lambda.$ Find the sum of the digits of $\lambda.$
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P162008
Yesterday at 12:04 AM
maromex
3 hours ago
Inequalities
sqing   27
N 4 hours ago by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
27 replies
sqing
May 13, 2025
sqing
4 hours ago
Divisors of factorials can't be always products of consecutive integers
Johann Peter Dirichlet   0
4 hours ago
Let $M$ an even number.

Show that $\frac{n!}{M^2}$ is not the product of consecutive integers for infinitely many naturals $n$.
0 replies
Johann Peter Dirichlet
4 hours ago
0 replies
IOQM P22 2024
SomeonecoolLovesMaths   3
N Yesterday at 10:51 PM by SomeonecoolLovesMaths
In a triangle $ABC$, $\angle BAC = 90^{\circ}$. Let $D$ be the point on $BC$ such that $AB + BD = AC + CD$. Suppose $BD : DC = 2:1$. if $\frac{AC}{AB} = \frac{m + \sqrt{p}}{n}$, Where $m,n$ are relatively prime positive integers and $p$ is a prime number, determine the value of $m+n+p$.
3 replies
SomeonecoolLovesMaths
Sep 8, 2024
SomeonecoolLovesMaths
Yesterday at 10:51 PM
AP calc?
Thayaden   30
N Yesterday at 9:53 PM by Pengu14
How are we all feeling on AP calc guys?
30 replies
Thayaden
May 20, 2025
Pengu14
Yesterday at 9:53 PM
Calculate the radius of a circle using sidelengths.
richminer   0
Yesterday at 6:17 PM
Given triangle ABC with incircle (I), with D being the touchpoint of (I) and BC. Let M be the tangent point of the A-Mixtilinear circle (internally tangent). A' is the reflection of A through I. Calculate the radius of the circle (MDA') using the side lengths of the triangle ABC.
0 replies
richminer
Yesterday at 6:17 PM
0 replies
Number of real roots
girishpimoli   0
Yesterday at 5:35 PM
Number of real roots of

$\displaystyle 2\sin(\theta)\cos(3\theta)\sin(5\theta)=-1$
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girishpimoli
Yesterday at 5:35 PM
0 replies
Factorization Ex.28a Q30
Obvious_Wind_1690   1
N Yesterday at 4:43 PM by Lankou
Please help with factorization. Given is the question


\begin{align*}
a(a+1)x^2+(a+b)xy-b(b-1)y^2\\
\end{align*}
And the given answer is


\begin{align*}
[(a+1)x-(b-1)y][ax+by]\\
\end{align*}
But I am unable to reach the answer.
1 reply
Obvious_Wind_1690
Yesterday at 4:17 AM
Lankou
Yesterday at 4:43 PM
Polynomials
P162008   4
N Yesterday at 4:19 PM by HAL9000sk
If $f(x)$ is a polynomial function such that $f(x) = x\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + \cdots}}}}$ then

A) Degree of $f(x)$ must be greater than $2$

B) $f(-2) = 0$

C) $\sum_{r=1}^{5} \frac{1}{f(r)} = \frac{25}{42}$

D) $\sum_{r=1}^{n} \frac{1}{f(r)} = \frac{n(3n + 5)}{4(n+1)(n+2)}$
4 replies
P162008
Monday at 11:18 PM
HAL9000sk
Yesterday at 4:19 PM
hard inequality
revol_ufiaw   10
N Yesterday at 3:43 PM by sqing
Prove that $(a-b)(b-c)(c-d)(d-a)+(a-c)^2 (b-d)^2\ge 0$ for rational $a, b, c, d$.
10 replies
revol_ufiaw
Yesterday at 1:09 PM
sqing
Yesterday at 3:43 PM
Pair of multiples
Jalil_Huseynov   64
N May 11, 2025 by Markas
Source: APMO 2022 P1
Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$.
64 replies
Jalil_Huseynov
May 17, 2022
Markas
May 11, 2025
Pair of multiples
G H J
Source: APMO 2022 P1
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Nguyenhuyhoang
207 posts
#51
Y by
Hi guys, can anyone check if the following solution is correct? Many thanks!

**Step 1**

Since $b-1$ is a multiple of $a-1$, we can write $b-1 = k(a-1)$ for some positive integer $k$. This implies
$$b \equiv 1 \pmod{a-1}.$$
**Step 2**

Since $a^3$ is a multiple of $b^2$, we have $b^2 \mid a^3$. By the divisor bounding inequality, we get
$$b \le a^{\frac{3}{2}}.$$
**Step 3**

Combining the results from Step 1 and Step 2, we have:
$$1 \le b \le a^{\frac{3}{2}} \text{ and } b \equiv 1 \pmod{a-1}.$$
This means $b$ must be one of the numbers $1, a, 2a-1, 3a-2, \dots$ up to the largest multiple of $a-1$ less than or equal to $a^{\frac{3}{2}}$.

**Step 4: Casework**

* **Case 1: b = 1**
This case trivially satisfies the conditions for any positive integer $a$.

* **Case 2: b = a**
This case also trivially satisfies the conditions for any positive integer $a$.

* **Case 3: b > a**
In this case, we have $b = l(a-1) + 1$ for some positive integer $l \ge 2$. Since $b^2 \mid a^3$, we can write
$$a^3 = m b^2 = m (l(a-1) + 1)^2$$for some positive integer $m$. Expanding the right side, we get a polynomial in $a$ with leading coefficient $ml^2$. Since the left side has leading coefficient $1$, we must have $ml^2 = 1$. This is impossible since $m$ and $l$ are both positive integers greater than 1.

**Step 5: Product of Two Numbers is a Square and Four Numbers Theorem**

We are left with the case where $a < b < 2a-1$. Since $b^2 \mid a^3$, we can write $a^3 = cb^2$ for some positive integer $c$. By the "Product of Two Numbers is a Square" technique, there exist positive integers $d, p, q$ with $\gcd(p,q) = 1$ such that
$$a = dp^2 \text{ and } c = dq^2.$$
Substituting into $a^3 = cb^2$, we get $d^3p^6 = d^2q^2b^2$, or $dp^6 = q^2b^2$. By the Four Numbers Theorem, there exist positive integers $r, s, t, u$ such that
$$d = rs, \ p^3 = tu, \ q = rt, \ b = su.$$
Since $\gcd(p,q) = 1$, we must have $\gcd(t,s) = 1$. Therefore, $s^2 \mid d$ and $t^2 \mid p^3$. This implies $s \mid r$ and $t \mid u$. Writing $r = vs$ and $u = wt$, we get
$$d = v s^2, \ p^3 = vwt^3, \ q = vst, \ b = su.$$
Substituting into $b-1 = k(a-1)$, we get
$$su - 1 = k(vs^2p^2 - 1) = k(vs^2vwt^3 - 1) = k(v^2s^2wt^3 - 1).$$
This implies $su - 1$ is divisible by $s^2$. However, since $s \ge 1$, we have $su - 1 \equiv -1 \pmod{s^2}$, a contradiction.

**Conclusion**

Therefore, the only solutions are $(a,b) = (a,1)$ and $(a,b) = (a,a)$ for any positive integer $a$.
This post has been edited 1 time. Last edited by Nguyenhuyhoang, May 30, 2024, 7:56 PM
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Ywgh1
139 posts
#53
Y by
apmo 2022 p1

Let $k=\frac{a^3}{b^2}$ then we have that $a-1| k-1$
Which only happens when $a=b=k$ and hence we get that our solutions are $(n,n)$ and $(n,1)$.
This post has been edited 1 time. Last edited by Ywgh1, Aug 27, 2024, 11:54 AM
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AshAuktober
1009 posts
#54
Y by
Case 1: $b<a \implies b = 1, a > 1$, which always works.
Case 2: $b = a$ also always works.
Case 3: $b>a$.
Let $a = dx, b = dy$ with $d = gcd(a, b)$.
The equations reduce to $$y^2 \mid d, dx-1 \mid x-y.$$Now let $d = ky^2$.
Then $$kxy^2 - 1 \mid x-y,$$so $$ky^2x-1 \le y-x \le xy-1$$$$\implies ky^2x \le xy \implies ky \le 1 \implies k = y = 1.$$But then $$b = dy = d \le dx = a,$$contradiction. So the only solutions are $\boxed{(a, b) = (t, 1), (t,t)}$ where $t \in \mathbb{N}$. $\square$
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SomeonesPenguin
129 posts
#55
Y by
Here is a quick solution. :-D

We will show that the only solutions are $(a,b)=(a,1)$ and $(a,b)=(a,a)$. If $b=1$, $a$ can be any positive integer (apart from $1$) so suppose that $b\neq 1$.

Let $a^3=k\cdot b^2$ where $k$ is a positive integer. Now look at this equation mod $a-1$ and keep in mind that $b\equiv 1 \pmod{a-1}$. We get $$k\equiv 1\pmod{a-1}$$
Now we begin our size argument.

Case 1. $k=1$

We have that $a^3=b^2=k^6$ so $a=k^2$ and $b=k^3$. Now we have $$k^2-1\mid k^3-1\iff k+1\mid k^2+k+1\iff k+1\mid 1$$
So we have no solutions.

Case 2. $k \ge a$

If $b=a$ we get a solution so suppose that $b\neq a$. We clearly have $a<b$ and $t=\frac{b-1}{a-1}\ge 2 \Rightarrow b \ge 2a-1$. Now we have $a^3=k\cdot b^2\ge a(2a-1)^2$. After expanding, this implies $$0\ge 3a^2-4a+1$$
And this is a contradiction. $\square$
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alexanderhamilton124
400 posts
#56
Y by
If $b = 1$, we are done, so assume not. Let $a^3 = kb^2$, we have $a - 1 \mid a^3 - 1 = kb^2 - 1$, $a - 1 \mid b - 1 \mid kb^2 - kb$, so $a - 1 \mid kb - 1 \implies a - 1 \mid (k - 1)b$. Since $a - 1 \mid b - 1$, $(a - 1, b) = 1 \implies a - 1 \mid k - 1$.

If $k = 1$, we have $a^3 = b^2$, so let $a = x^2$, $b = x^3$. $x^2 - 1 \mid x^3 - 1 \implies x + 1 \mid x^2 + x + 1$, a contradiction.

If $k \geq a$, then $kb^2 > a^3$, unless $b = k = a$, which works, so we're done.
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RedFireTruck
4243 posts
#57
Y by
Clearly $b=1$ works, so assume $b>1$.

Let $a^3=nb^2$ and $n=c^3d$ where $n,c,d\ge 1$ are integers and $c$ is maximized. Also note that $n\le b$ because $a\le b$. We see that we can let $a=cdk^2$ and $b=dk^3$ where $k\ge 1$ is an integer. Since $n\le b$, this means that $c\le k$. Plugging this in gives $(cdk^2-1)|(dk^3-1)$ so $(cdk^2-1)|(cdk^3-c)$ so $(cdk^2-1)|(k-c)$. Plugging in $k=c$ gives $a=b$, which works. When $c < k$, the LHS is bigger than the RHS.

Therefore, the solutions are $\boxed{b=1}$ and $\boxed{a=b}$.
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Sadigly
229 posts
#58 • 1 Y
Y by alexanderhamilton124
Solution by Mhremath & Sadigly

$a-1\mid b-1$ Either $b=1$ or $b\geq a$. For $b=1$ case, $(a;b)=(a;1)$ obviously works

Let $a^3=kb^2$ for a $k\in\mathbb{Z}^+$

Claim. $a-1\mid k-1$
Proof. We have $a-1\mid (b-1)(b+1)k$ or in other words $a-1\mid b^2k-k\Rightarrow a-1\mid a^3-k$. We also have $a-1\mid a^3-1$ for $a\neq1$. Subtracting these two gives us $a-1\mid k-1$

This could mean 3 things.

1. Case $k=1\Rightarrow a^3=b^2=x^6$ for some $x\in\mathbb{Z}^+/\{1\}$
$$a-1\mid b-1\Rightarrow x^2-1\mid x^3-1\Rightarrow x+1\mid x^2+x+1$$But we have $gcd(x+1;x^2+x+1)=1$ So this case can't be true

2. Case $k=a\Rightarrow a=b$ This obviously works.

3. Case $k>a\Rightarrow a^3=kb^2>ab^2\Rightarrow a>b$ But we have $b\geq a$, so this case can't be true,too
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math004
23 posts
#59
Y by
Let $k=a^3/b^2.$
\[0\equiv a^3-1 = kb^2-1\equiv k-1 \pmod{a-1}.\]However, $a-1\mid b-1\implies b\leq a \implies k\leq a.$ This forces $k=a,$ in which case $b^2=a^2\implies a=b$ or $k=1.$
When $k=1, a^3=b^2\implies (a,b)=(x^2,x^3).$ $$a-1\mid b-1 \iff x^2-1\mid x^3-1  \iff x+1\mid 1+x+x^2 \iff x+1\mid 1.$$Which is obviously impossible. Convsersely, We can easily verify that $(a,a)$ is a solution.
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pie854
243 posts
#60
Y by
Taking $b=1$ works. Suppose $b>1$, then $a-1\mid b-1$ gives $a\leq b$. Note that $b^2(a-1) \mid a^3(b^2-1)$. So, $$b^2a-b^2 \mid -a^3(b^2-1)+(a^2+a+1)(b^2a-b^2)=a^3-b^2.$$Thus, $b^2a-b^2\leq a^3-b^2 \implies b\leq a$ (if $a^3=b^2$ then let $a=x^2$, $b=x^3$ and thus $x^2-1\mid x^3-1$ i.e. $x=1$, a contradiction) and it follows that $a=b$.
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SSS_123
20 posts
#61
Y by
Given conditions,
$b^2|a^3$
$a-1|b-1$
where $a,b$ are positive integers.
Let $a,b \neq 1$
From the 2nd condition we have,
$a-1 \leq b-1$
$a \leq b$
From the first condition we have,
$b^2|a^3$
So we can write $a^3=b^2k$ for some positive integer $k$.

Now
$ b-1 \equiv 0 \mod{(a-1)}$
$ b \equiv 1 \mod{(a-1)}$
$ b^2 \equiv 1 \mod{(a-1)}$
$ b^2k \equiv k \mod{(a-1)}$
$ a^3 \equiv k \mod{(a-1)}$
$ 1 \equiv k \mod{(a-1)}$
$ 1  \equiv \frac{a^3}{b^2}  \mod{(a-1)}$
$ \frac{a^3}{b^2} - 1 \equiv 0 \mod{(a-1)}$
So
$a-1|\frac{a^3}{b^2} - 1$
$a-1|\frac{a^3-b^2}{b^2}$
$b^2(a-1)|a^3-b^2$
$ab^2-b^2|a^3-b^2$
$ab^2-b^2 \leq a^3-b^2$
$ab^2 \leq a^3$
$b^2 \leq a^2$
$b \leq a$
But we also got $a \leq b$
Thus $a=b$
Now if $a=1$ then it is obvious that the conditions won't fullfill
if $b=1$,it is easy to see that it works for all positive integers $a$
Thus the required solutions are:
$(a,b)=(x,1),(x,x)$ for all positive integers $x$.
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Iveela
117 posts
#62
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Answer: $(a, b) = (x, 1)$ and $(x, x)$.

Suppose there exist other solutions. Let $b - 1 = k(a - 1)$ which we can rewrite as $b = (a - 1)k + 1$. Then $d = \text{gcd}(a, b) = \text{gcd}(a,  k - 1) \leq k - 1$. Now, notice that $b^2 \mid a^3 \Leftrightarrow \left (\frac{b}{d} \right)^2 \mid \left( \frac{a}{d} \right)^3 \cdot d$ implies $\left( \frac{b}{d} \right)^2 \mid d$. Consequently, we get the following bound for $k$.
\[\frac{b^2}{d^2} \leq d \implies  a^2 \leq b^2 \leq d^3 \leq (k - 1)^3 \implies a^{\frac{2}{3}} + 1 \leq k.\]Plugging this back in, we receive $b \geq (a - 1)(a^{\frac{2}{3}} + 1) + 1 \geq a^{\frac{5}{3}}$ which implies $b^2 > a^3$, a contradiction. $\square$
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ray66
46 posts
#63
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$(a,b)=(x,1)$ and $(x,x)$ for a positive integer $x$.

First, verify that $b=1$ works. Next, assume $b \ge 2$. Then $b \ge a$ from $b-1$ is a multiple of $a-1$. Also, $a^3=qb^2$. If $q=a$, then $a=b$. Otherwise, $b^2 \equiv 0 \pmod a$ so $a | b$. Writing $b-1 = k(a-1)$ gives $$ b-1 \equiv -k \pmod a$$or $$k \equiv 1 \pmod a$$. $k=a+1$ gives $b=a^2$, but then $a^4 | a^3$, contradiction. So $k=1$ and $a=b$.
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Ilikeminecraft
664 posts
#64
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By the first condition, we have that $a^3 = kb^2$ for some $k\in\mathbb Z.$ Hence, $a - 1 \mid b - 1 \mid b^2 - 1 \mid kb^2 - k = a^3 - k.$ Hence, we conclude $k\equiv 1\pmod {a - 1}.$ If $k\neq a,$ then $k\geq 2a - 1,$ which is clearly absurd since our second condition implies $b \geq a.$ Thus, $k = a,$ and so $\boxed{a = b > 1}$ is the only solution.
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NerdyNashville
18 posts
#65
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Solution
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Markas
150 posts
#66
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Let $a^3 = k.b^2$. Now $a-1 \mid a^3 - 1 = k.b^2 - 1$. Also $a-1 \mid kb(b-1)$ $\Rightarrow$ $a-1 \mid k.b^2 - 1 - kb.(b-1) = kb - 1$, but $a - 1 \mid k(b-1)$ $\Rightarrow$ $a - 1 \mid kb - 1 - k(b-1) = k - 1$ $\Rightarrow$ $a - 1 \mid k - 1$ and $a - 1 \mid b - 1$ $\Rightarrow$ $a \leq k$ and $a \leq b$. Now $a^3 = k.b^2 \geq a.a^2 = a^3$ $\Rightarrow$ a = b $\Rightarrow$ the pairs that work are (a,b) = (a,1); (a,a) and we are ready.
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