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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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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
13 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.
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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
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
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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 $n$.
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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 $m+n+p$.
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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?
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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.
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richminer
Yesterday at 6:17 PM
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Number of real roots
girishpimoli   0
Yesterday at 5:35 PM
Number of real roots of

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girishpimoli
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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) $\sum_{r=1}^{n} \frac{1}{f(r)} = \frac{n(3n + 5)}{4(n+1)(n+2)}$
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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 $a, b, c, d$.
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revol_ufiaw
Yesterday at 1:09 PM
sqing
Yesterday at 3:43 PM
Beautiful Number Theory
tastymath75025   35
N Monday at 7:42 AM by N3bula
Source: 2022 ISL N8
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
35 replies
tastymath75025
Jul 9, 2023
N3bula
Monday at 7:42 AM
Beautiful Number Theory
G H J
G H BBookmark kLocked kLocked NReply
Source: 2022 ISL N8
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tastymath75025
3223 posts
#1 • 3 Y
Y by Rounak_iitr, tofubear, cubres
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
This post has been edited 1 time. Last edited by tastymath75025, Jul 9, 2023, 4:20 PM
Reason: fix wording
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tastymath75025
3223 posts
#2 • 3 Y
Y by tofubear, cubres, Kingsbane2139
Note that if $n$ is even then $3\mid2^n+65$ but $3\not | 5^n-3^n$, impossible, so $n$ is odd, and clearly $n=1$ fails. Now we claim all $n>1$ fail. To see why, note that

\[-1 = \left( \frac{2^n+65}{5} \right) = \left( \frac{5}{2^n+65} \right) = \left( \frac{5^n}{2^n+65} \right),\]
and now if $2^n+65 \mid 5^n-3^n$ this in turn equals

\[\left( \frac{3^n}{2^n+65} \right) = \left( \frac{3}{2^n+65} \right) = \left( \frac{2^n+65}{3} \right) = +1,\]
a contradiction.

(Here we make use Jacobi symbols.)
This post has been edited 1 time. Last edited by tastymath75025, Jul 9, 2023, 4:49 AM
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DottedCaculator
7357 posts
#3 • 3 Y
Y by centslordm, ehuseyinyigit, cubres
1998 n5

Suppose $2^n+65\mid5^n-3^n$. Since $3\nmid 5^n-3^n$ and $5\nmid 5^n-3^n$, we get $n\equiv1\pmod2$. Therefore, $5^n-3^n=5x^2-3y^2$ for some $x$ and $y$, implying $15\equiv\frac{(3y)^2}{x^2}\pmod{2^n+65}$, so $\left(\frac{15}{2^n+65}\right)=1$. Clearly, $2^1+65\nmid5^1-3^1$, so $n\geq2$. However, we also have
\begin{align*}
\left(\frac{15}{2^n+65}\right)&=\left(\frac{2^n+65}{15}\right)\\
&=\left(\frac{2^n+65}5\right)\left(\frac{2^n+65}3\right)\\
&=\left(\frac{2^n}5\right)\left(\frac{2^n+2}3\right)\\
&=(-1)(1)\\
&=-1,
\end{align*}contradiction. Therefore, $2^n+65\nmid 5^n-3^n$ for all $n$.
This post has been edited 1 time. Last edited by DottedCaculator, Jul 20, 2023, 10:18 AM
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brainfertilzer
1831 posts
#4 • 2 Y
Y by hdnlz, cubres
solution
This post has been edited 1 time. Last edited by brainfertilzer, Jul 9, 2023, 5:48 PM
Reason: n is positive, not nonnegative
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VicKmath7
1391 posts
#5 • 2 Y
Y by ehuseyinyigit, cubres
Quite similar to Romania TST 2008/3/3 and exactly the same approach works here. Definitely easier than N5 and N6.

Solution of N8
This post has been edited 2 times. Last edited by VicKmath7, Jul 10, 2023, 8:14 AM
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MatteD
90 posts
#6 • 1 Y
Y by cubres
Isn't the wording supposed to be this one?
Quote:
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
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IAmTheHazard
5003 posts
#7 • 2 Y
Y by centslordm, cubres
No such $n$. By taking modulo $3$, $n$ is odd, and it is clear that $n=1$ fails, so assume $n \geq 2$. Therefore, if $(\tfrac{5}{3})^n-1 \equiv 0 \pmod{2^n+65}$, $\tfrac{5}{3}$ must be a quadratic residue modulo $2^n+65$. However, letting $(\tfrac{a}{b})$ denote the Jacobi symbol, by quadratic reciprocity we have
$$\left(\frac{5/3}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)\left(\frac{2^n+65}{3}\right)=\left(\frac{\pm 2}{5}\right)\left(\frac{1}{3}\right)=(-1)(1)=-1,$$contradiction. $\blacksquare$

Remark: "IT'S [redacted] JACOBI" [dies]
This post has been edited 1 time. Last edited by IAmTheHazard, Jul 9, 2023, 12:42 PM
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blackmetalmusic
30 posts
#8 • 4 Y
Y by Kingsbane2139, hdnlz, ehuseyinyigit, cubres
Is this really n8?
It's just some simple quadratic residue stuff
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megarnie
5611 posts
#9 • 3 Y
Y by GeorgeRP, ehuseyinyigit, cubres
Definition: In the solution below, $\left( \frac{a}{b} \right)$ denotes the Jacobi symbol.

No solutions.

Since $n = 1$ fails, assume that $n > 1$. Note that $2^n + 65 \equiv 1\pmod 4$.

Clearly $n$ even fails, as $3$ would divide $2^n + 65$ but $3$ cannot divide $5^n - 3^n$. Now assume $n$ is odd.

Notice that $\frac{\left( \frac{5^n}{2^n + 65} \right)}{ \left( \frac{3^n}{2^n + 65} \right) }$ is equal to $1$ (since $5^n\equiv 3^n \pmod{2^n + 65}$), since $n$ is odd, this implies $\frac{\left( \frac{5}{2^n + 65} \right)}{ \left( \frac{3}{2^n + 65} \right) } = 1$.


However, we have that \[\left( \frac{5}{2^n + 65} \right) = \left( \frac{2^n + 65}{5} \right) = -1\]and \[\left( \frac{3}{2^n + 65} \right) = \left( \frac{2^n + 65}{3} \right) = \left( \frac{1}{3}  \right) = 1,\]absurd.

Note: Another way to show that $\frac{\left( \frac{5}{2^n + 65} \right)}{ \left( \frac{3}{2^n + 65} \right) } = 1$ is to let $p$ be a prime dividing $2^n + 65$. Then $1 = \left( \frac{ \frac{5}{3} }{p}  \right) = \frac{ \left( \frac{5}{p} \right) }{ \left( \frac{3}{p} \right)} $. Since this is true for all primes $p\mid 2^n + 65$, we have the desired result.
This post has been edited 5 times. Last edited by megarnie, Jul 9, 2023, 6:48 PM
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Leo.Euler
577 posts
#10 • 1 Y
Y by cubres
Why are N8s easier than N5s?

I claim the answer is $n=0$. Assume for contradiction that there exists a solution for $n > 0$.

Taking $\pmod{3}$ of the fractional expression, it is clear that $n$ must be odd (because otherwise $3$ divides the denominator but not the numerator). Let $(\tfrac{a}{b})$ denote the Jacobi symbol henceforth.

Note that $(5/3)^n \equiv 1 \pmod{2^n+65}$, and as $n$ is odd, it is clear that $5/3$ must be a quadratic residue (mod $2^n+65$). But:

\begin{align*}
\left(\frac{5/3}{2^n+65}\right) &= \left(\frac{5}{2^n+65}\right) \left(\frac{3^{-1}}{2^n+65}\right) \\
&= \left(\frac{5}{2^n+65}\right) \left(\frac{3}{2^n+65}\right)
&= \left(\frac{2^n+65}{3}\right) \left(\frac{2^n+65}{5}\right)
&= 1 \cdot \left(\frac{2}{5}\right)^n
&= 1 \cdot (-1)
&= -1,
\end{align*}a contradiction, and we conclude. $\blacksquare$
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cj13609517288
1924 posts
#11 • 1 Y
Y by cubres
lol what

Assume FTSOC that there does exist some $n$ such that $2^n+65\mid 5^n-3^n$. If $n$ is even, then $2^n+65$ is a multiple of $3$, impossible. So $n$ is odd.

Let $p$ be a prime dividing $2^n+65$. Obviously $p\not\in\{2,3,5\}$. Then
\[p\mid 2^n+65\mid 5^n-3^n\mid 15^n-9^n,\]so $15$ is a QR mod $p$.

By Quadratic Reciprocity,
\[\left(\frac{p}{3}\right)\left(\frac{p}{5}\right)=\left(\frac{15}{p}\right)\left(\frac{p}{15}\right)=(-1)^{(p-1)/2}.\]Caseworking, we get that $p$ mod $60$ is one of $S=\{1,7,11,17,43,49,53,59\}$. Under further inspection, we note that $S$ is closed under multiplication mod $60$, so $2^n+65$ mod $60$ is also in $S$. So $2^n$ mod $60$ is in
\[\{2,6,12,38,44,48,54,56\}.\]Since it must be a multiple of $4$ and not a multiple of $3$, it must be one of $\{44,56\}$. But that means $2^n$ is $1$ or $4$ mod $5$, impossible since $n$ is odd. $\blacksquare$
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BlazingMuddy
282 posts
#12 • 3 Y
Y by sabkx, pavel kozlov, cubres
May I give a small comment to the placement of this problem?

It seems that this problem is a kind of a problem that gets killed with one trick... Which reminds me of ISL 2017 N8. The difficulty in finding the trick is supposed to be what makes the problem an N8. The difference between this thing and ISL 2017 N8, in my opinion, is that the kind of trick that kills ISL 2017 N8 is something that might be too hard to find in a contest setting. Meanwhile this? I'm not so sure about that...

Anyways, let's replace $5$ and $65$ with arbitrary odd positive integers, say $A$ and $B$, and see how well this problem generalizes. For $n$ even, you want $3 \mid 2^n + B$, so $B \equiv 2 \pmod{3}$; also you want $3 \nmid A$. For $n = 1$, you just want $B + 2 \nmid A - 3$. For $n$ odd with $n > 1$, applying Jacobi symbol, you want that
$$ \left(\frac{A}{2^n + B}\right) = -\left(\frac{3}{2^n + B}\right) = (-1)^{(B + 1)/2} \left(\frac{2^n + B}{3}\right) = (-1)^{(B + 1)/2}. $$For a huge convenience, you also want $A \mid B$, so
$$ \left(\frac{A}{2^n + B}\right) = (-1)^{(A - 1)(B - 1)/4} \left(\frac{2}{A}\right). $$We can now pick $B \equiv 1 \pmod{4}$ and $A \equiv 3, 5 \pmod{8}$ (as in the original problem). In summary...
Quote:
If $A \equiv 3, 5 \pmod{8}$, $3 \nmid A$, $B \equiv 5 \pmod{12}$, $A \mid B$, and $B + 2 \nmid A - 3$, then $2^n + B$ does not divide $A^n - 3^n$ for any $n \geq 1$.
Quote:
If $A \equiv 3, 5 \pmod{8}$, $3 \nmid A$, $B \equiv 5 \pmod{12}$, and $A \mid B$, then $2^n + B$ does not divide $A^n - 3^n$ for any $n > 1$.
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Orthogonal.
594 posts
#13 • 1 Y
Y by cubres
Huh.

Assume that there exists an $n$ such that $\frac{5^n-3^n}{2^n+65}$ is an integer. By taking mod $3$, $n$ must be odd. Clearly, $n=1$ fails. Let $\left (\frac{a}{b}\right)$ denote the Jacobi symbol. We then have that $$\frac{\left(\frac{5^n}{2^n+65}\right)}{\left(\frac{3^n}{2^n+65}\right)}=\frac{\left(\frac{5}{2^n+65}\right)}{\left(\frac{3}{2^n+65}\right)}=1.$$
But $$\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right) = 1$$and $$\left(\frac{5}{2^n+65}\right) = \left(\frac{2^n+65}{5}\right) = -1,$$
a contradiction.
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Sourorange
107 posts
#14 • 1 Y
Y by cubres
Is this really difficult enough to be N8?
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vsamc
3789 posts
#15 • 2 Y
Y by kamatadu, cubres
Solution
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awesomeming327.
1736 posts
#16 • 1 Y
Y by cubres
We use the Jacobi symbol. Taking $\pmod 3$ forces $n$ odd. We have for $n\ge 3$
\[\left(\frac{5\cdot 3^{-1}}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)\left(\frac{2^n+65}{3}\right)=(-1)(1)=-1\]a contradiction because we need $5\cdot 3^{-1}\equiv 1\pmod {2^n+65}$.
This post has been edited 1 time. Last edited by awesomeming327., Aug 4, 2023, 5:35 PM
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Sourorange
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#17 • 1 Y
Y by cubres
Here is the official answer proposed by IMO committee.
Attachments:
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TheBigMath13876
2 posts
#18 • 1 Y
Y by cubres
I've just started pursuing math at this level, so I'm not at all good at all of this. With that being said, does anyone know if there is another way to solve this without the use of modules?
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Hertz
32 posts
#20 • 1 Y
Y by cubres
Let's assume $n$ is even, then $3 \mid 2^n + 65$ but $3 \nmid 5^n - 3^n$ contradiction. Since $n \neq 1$ we get $n \ge 3$ odd number.

Now let's assume that $2^n + 65 \mid 5^n-3^n$ then we get $5^n \equiv 3^n ($mod $ 2^n+65)$ Now let us introduce the jaccobi symbol. We get

$$\left( \frac{5^n}{2^n+65} \right) = \left (\frac{3^n}{2^n+65} \right)$$
Now lets count them seperatly:
$$\left( \frac{3^n}{2^n+65} \right) = \prod_{i=1}^n \left( \frac{3}{2^n+65} \right)^n = \left( \frac{3}{2^n+65} \right) $$
But since $2^n + 65 \equiv 1 ($mod $ 4) $ we get $\left( \frac{3}{2^n+65} \right) = \left( \frac{2^n+65}{3} \right) = 1 $

$$\left( \frac{5^n}{2^n+65} \right) = \prod_{i=1}^n \left( \frac{5}{2^n+65} \right)^n = \left( \frac{5}{2^n+65} \right) $$
Since $5 \equiv 1 ($mod $  4)$ we get $\left( \frac{5}{2^n+65} \right) = \left( \frac{2^n+65}{5} \right)$

But $2^n \equiv 2;3 ($mod $ 5)$ and $5 \mid 65$ so $\left( \frac{2^n+65}{5} \right) = -1$ a contradiction
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hectorleo123
347 posts
#21 • 1 Y
Y by cubres
tastymath75025 wrote:
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
Here $\left(\frac{a}{b}\right)$ is the Jacobi symbol
Suppose there exists $n$ such that
$$2^n+65|5^n-3^n$$If $n=1\Rightarrow 67|2(\Rightarrow \Leftarrow)$
$$\Rightarrow n>1$$$\color{red}\boxed{\textbf{If n is even:}}$
$\color{red}\rule{24cm}{0.3pt}$
$$\Rightarrow 2^n\equiv 1\pmod{3}$$$$\Rightarrow 2^n+65\equiv 0\pmod{3}$$$$\Rightarrow 3|5^n-3^n$$$$\Rightarrow 3|5^n(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{If n is odd:}}$
$\color{red}\rule{24cm}{0.3pt}$
$$2^n+65|5^n-3^n$$$$\Rightarrow \left( \frac{5^n}{2^n+65} \right)=\left( \frac{3^n}{2^n+65} \right)$$$$\Rightarrow \left( \frac{5}{2^n+65} \right)^n=\left( \frac{3}{2^n+65} \right)^n$$$$\Rightarrow \left( \frac{5}{2^n+65} \right)=\left( \frac{3}{2^n+65} \right)$$$$\Rightarrow \left( \frac{2^n+65}{5} \right)(-1)^{\frac{(2^n+64)(4)}{4}}=\left( \frac{2^n+65}{3} \right)(-1)^{\frac{(2^n+64)(2)}{4}}$$Since $n>1\Rightarrow \frac{(2^n+64)(2)}{4}$ is even:
$$\Rightarrow \left( \frac{2^n}{5} \right)=\left( \frac{2^n-1}{3} \right)$$$$\Rightarrow \left( \frac{2(2^{n-1})}{5} \right)=\left( \frac{1}{3} \right)=1$$Since $n$ is odd $\Rightarrow 2^{n-1}$ is a perfect square
$$\Rightarrow \left( \frac{2}{5} \right)=1$$$$\Rightarrow -1=1(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$
$$\Rightarrow 2^n+65\nmid 5^n-3^n$$$\color{blue}\rule{24cm}{0.3pt}$
This post has been edited 2 times. Last edited by hectorleo123, Mar 13, 2024, 3:27 PM
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M.rastgar
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#22 • 1 Y
Y by cubres
pretty same as #2. n is odd . Let k= 2^n +65 then we have 5^n = 3^n mod k then (15/k)=1 this means (3/k)= (5/k) but we have (3/k) = 1 and (5/k) = -1 which is contradiction.
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math_comb01
662 posts
#23 • 1 Y
Y by cubres
Jacobi kills it, probably same as others.
ISL 2022 N8 wrote:
Prove that $2^n+65$ does not divide $5^n-3^n$ for any positive integer $n$
Clearly $n$ has to be odd, because if it is even then $3 \mid 2^n+65 \mid 5^n-3^n$ which is a contradiction, $5^n \equiv 3^n (\mod 2^n+65)$
then:
$\left(\frac{5^n}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)=-1$
$\left(\frac{3^n}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=1$
where the last 2 lines are in jacobi symbol. Contradiction...
This post has been edited 5 times. Last edited by math_comb01, Mar 21, 2024, 3:15 PM
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thdnder
198 posts
#24 • 1 Y
Y by cubres
Assume $2^n + 65 \mid 5^n - 3^n$. If $n \equiv 0 (2)$, then $3 \mid 2^n + 65 \mid 5^n - 3^n$, a contradiction. Thus $n \equiv 1 (2)$. Now note that $\left(\frac{5^n}{2^n + 65} \right) = \left(\frac{3^n}{2^n + 65}\right)$. By using quadratic reciprocity law, we get $\left(\frac{3^n}{2^n + 65}\right) = \left(\frac{2^n + 65}{3^n}\right) = \left(\frac{2^n + 65}{3}\right)^n = \left(\frac{2^n + 65}{3}\right) = \left(\frac{1}{3}\right) = 1$.

But $\left(\frac{5^n}{2^n + 65}\right) = \left(\frac{2^n + 65}{5^n}\right) = \left(\frac{2^n + 65}{5}\right) = \left(\frac{2^n}{5}\right) = \left(\frac{2}{5}\right) = -1$, which is an evident contradiction. $\blacksquare$
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ihategeo_1969
242 posts
#25 • 1 Y
Y by cubres
Assume some $n$ works.

Obviously $n=1$ does not work for size reasons. And also even $n$ does not work because we get $3 \mid 2^n+65 \mid 5^n-3^n \implies 3 \mid 5$, which I think is wrong as far as I know.

See that we get \[\left(\frac{5^n}{2^n+65}\right)=\left(\frac{3^n}{2^n+65}\right) \iff \left(\frac{5}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right) \iff \left(\frac{2^n+65}5\right)=\left(\frac{2^n+65}3 \right) \iff \left(\frac{2^n}5 \right)=\left(\frac{2^n-1}3\right)\]which is obviously false for odd $n$ because it implies $\left(\frac{2^n}5\right)=-1$ but $\left(\frac{2^n-1}3\right)=1$, a contradiction.
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OronSH
1748 posts
#26 • 2 Y
Y by Clew28, cubres
First $n\ne 1$ clearly and $n\not\equiv 0\pmod 2$ by $\pmod 3.$ Now if $p\mid 5^n-3^n$ then $(5/3)^n\equiv 1\pmod n$ with $n$ odd, so $\left(\frac{5/3}p\right)=\left(\frac{15}p\right)=1.$ Now we must have $\left(\frac{15}{2^n+65}\right)=1,$ which is equivalent by QR to $\left(\frac{2^n+5}{15}\right)=1.$ But $2^n\equiv 2,8\pmod{15}$ for odd $n,$ and we can check $\left(\frac7{15}\right)=\left(\frac{13}{15}\right)=-1,$ contradiction.
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AshAuktober
1009 posts
#27 • 1 Y
Y by cubres
What's the motivation behind using the Jacobi symbol here?
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Vulch
2709 posts
#28 • 1 Y
Y by cubres
Can anyone solve it without Jacobi symbol?
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Mr.Aura
1 post
#29 • 2 Y
Y by Nobitasolvesproblems1979, cubres
Clearly when $n$ is even we get a contradiction, so $n$ is odd and $n$ $\geq$ 3.

Now, using Jacobi's symbol we get

$\left(\frac{5^n}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)=-1$, and

$\left(\frac{3^n}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=1$.

But this two must be equal, then we have a contradiction.
This post has been edited 2 times. Last edited by Mr.Aura, Dec 13, 2024, 12:43 PM
Reason: I forgot to write
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Eka01
204 posts
#30 • 2 Y
Y by cubres, L13832
Solved on the recommendation of the orz Sammy27.
If $n$ is even then $A=2^n +65$ is divisible by $3$ which doesn't divide $5^n-3^n$.
Henceforth assume $n$ is odd. This implies $5^n \equiv 3^n(mod \ A) \implies \left( \frac{5^n}{A} \right) =\left( \frac{3^n}{A} \right)$.
Now since the jacobi function is multiplicative, we have $\left( \frac{5^n}{A} \right) = \left( \frac{5}{A} \right)= \left( \frac{A}{5} \right)=-1$
And $\left( \frac{3^n}{A} \right) = \left( \frac{3}{A} \right)= \left( \frac{A}{3} \right)=1$
Which is a contradiction, hence there's no solutions for any $n$.
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orangesyrup
130 posts
#31 • 1 Y
Y by cubres
any sols without using jacobi?
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zuat.e
68 posts
#32 • 1 Y
Y by cubres
Claim: $n$ is odd
Note that $3\nmid 5^n-3^n$, hence $3 \nmid 2^n + 65$. As:
\[\left\{\begin{array}{cl}
             2^n \equiv 1&n \mbox{ odd}  \\
             2^n \equiv 2&n \mbox{ even} 
        \end{array}\right.\]$2^n+65\equiv 2^n+2 \equiv \{0,1\} \pmod{3}$ and for $2^n+65\equiv 1\pmod{3}$, we must have $n$ odd.
Therefore, looking at $\pmod{5}$, we have:
\[\left\{\begin{array}{cc}
         2^n\equiv2&n\equiv1\pmod{4}  \\
         2^n\equiv4&n\equiv2\pmod{4}  \\
         2^n\equiv3&n\equiv3\pmod{4}  \\
         2^n\equiv1&n\equiv4\pmod{4}  \\
    \end{array}\right.\]and as $n$ is odd, $2^n \equiv 2,3 \pmod{5}$
We now rewrite $(\frac{5}{3})^n\equiv1 \pmod{2^n+65}$, consequently $(\frac{5}{3})^{n+1}\equiv \frac{5}{3} \pmod{2^n+65}$, implying $\frac{5}{3}$ is $QR$ $\pmod{2^n + 65}$.
Nonetheless:\[\left( \frac{(\frac{5}{3})}{2^n +65}\right) \equiv \left( \frac{5}{3}\right)\left( \frac{5}{2^n+65}\right)\equiv \left( \frac{2^n+65}{5}\right)\equiv\left( \frac{\{2,3\}}{5}\right)\]but: \[\left( \frac{2}{5}\right), \left( \frac{3}{5}\right)\equiv -1\]therefore as: \[\left( \frac{(\frac{5}{3})}{2^n +65})\right)=\prod_{p\mid 2^n +65}\left(\frac{(\frac{5}{3})}{p}\right)\]there exists $p\mid 2^n +65$ prime such that $(\frac{5}{3})$ isn't $QR$ $\pmod{p}$, contradicting the fact that $(\frac{5}{3})$ is $QR$ $\pmod{2^n +65}$
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cubres
119 posts
#33
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Storage - grinding ISL problems
This post has been edited 2 times. Last edited by cubres, Jan 29, 2025, 9:03 PM
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Ilikeminecraft
664 posts
#34
Y by
triv n8???
the answer is no solutions
If $n$ is even, then $3\mid65 + 2^n\mid 5^n-3^n,$ contradiction
Thus, $n$ is odd. Specifically, we can get that $5^{2k + 1} \equiv 3^{2k +1}\pmod{2^n + 65}.$ This implies $15$ is a quadratic residue modulo $2^n + 65.$ However,
\begin{align*}
\left(\frac{15}{2^n+65}\right) & = \left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right) \\
& = \left(\frac{5}{2^n}\right) = -1
\end{align*}which is a contradiction.
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awesomehuman
499 posts
#35
Y by
Assume toward a contradiction $n$ works.

Claim: $n$ is odd.
Proof: Assume toward a contradiction $n$ is even. Then, $3\mid 2^n+65$, but $3\nmid 5^n-3^n$. $\blacksquare$

Let $p$ be a prime factor of $2^n+65$. Then,
\[\left(\frac{5}{p}\right) = \left(\frac{5^n}{p}\right) = \left(\frac{3^n}{p}\right) = \left(\frac{3}{p}\right).\]
Let $a$ and $b$ be positive integers such that $ab = 2^n+65$ and every prime factor of $a$ is $1\pmod{4}$ and every prime factor of $b$ is $3\pmod{4}$.

Let $p\mid a$. Then, by quadratic reciprocity, $\left(\frac{p}{5}\right) = \left(\frac{5}{p}\right) = \left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)$.

Let $p\mid b$. Then, by quadratic reciprocity, $\left(\frac{q}{5}\right) = \left(\frac{5}{q}\right) = \left(\frac{3}{q}\right) = -\left(\frac{q}{3}\right)$.

Let $b=p_1\dots p_k$ with $p_1,\dots, p_k$ prime.
Because $1\equiv 2^n+65\equiv ab\equiv b \equiv p_1\dots p_k \equiv 3^k\pmod{4}$, $k$ is even.
So, we have
\[\left(\frac{b}{5}\right) = \prod_{i=1}^k \left(\frac{p_i}{5}\right) = \prod_{i=1}^k -\left(\frac{p_i}{3}\right) =  \prod_{i=1}^k \left(\frac{p_i}{3}\right) = \left(\frac{b}{3}\right).\]By similar logic, $\left(\frac{a}{5}\right) = \left(\frac{a}{3}\right)$. Multiplying, we get $\left(\frac{2^n+65}{5}\right) = \left(\frac{2^n+65}{3}\right)$.

However, since $n$ is odd, the LHS is $-1$ and the RHS is $1$, a contradiction.
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Adywastaken
64 posts
#36
Y by
$2^n+65\equiv 0 \pmod 3 \implies n=2k+1$.
Let $3^k=v$, $5^k=u$.
$\left(\frac{5v}{u}\right)^2 \equiv 15 \pmod {2^n+65}$.
For odd n,

\[
\left(\frac{15}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right) \left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right) \left(\frac{2^n+65}{5}\right)=\left(\frac{2^n+2}{3}\right) \left(\frac{2^n}{5}\right)=(-1)(1)=-1
\]
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N3bula
296 posts
#37
Y by
$\color{blue} \boxed{\textbf{Claim: }n\textbf{ must be odd}}$

Proof:
Suppose $n$ is even, and take both expressions modulo $3$.
\[(1) \quad 2^n+65\equiv 1 +65\equiv 66\equiv 0 \pmod 3\]\[(2) \quad 5^n-3^n\equiv 5^n\not\equiv 0\pmod 3\]Thus combining $(1)$ and $(2)$ yields a contradiction.

$\color{blue} \boxed{\textbf{Claim: }n\textbf{ cannot be odd}}$

Proof:
If we have a prime $p$ such that $p\mid 2^n+65$ such that:
\[\left(\frac{5}{p}\right)\neq \left(\frac{3}{p}\right)\]We get a contradiction as if $p\mid 2^n+65$ and $2^n+65\mid 5^n-3^n$ we have that:
\[5^n\equiv 3^n\pmod p\]so we get:
\[\left(\frac{5^n}{p}\right)= \left(\frac{3^n}{p}\right)\]as $n$ is odd this means:
\[\left(\frac{5}{p}\right)= \left(\frac{3}{p}\right)\]Which is a contradiction. Now I will show such a $p$ exists. Using the jacobi symbol we get:
\[(1)\quad \left(\frac{5}{2^n+65}\right) = \left(\frac{2^n+65}{5}\right)=1\]\[(2)\quad \left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=-1\]These follow because $n$ is odd and $2^n+65\equiv 1\pmod 4$. Now let $2^n+65=p_1^{\alpha_1}\dots p_k^{\alpha_k}$.
The definition of the jacobi symbol gives us that:
\[\left(\frac{k}{2^n+65}\right)=\prod_{i=1}^{k}\left(\frac{k}{p_i}\right)^{\alpha_i}\]If for all such $p_i$ we have that
\[\left(\frac{5}{p_i}\right)= \left(\frac{3}{p_i}\right)\]we get that:
\[\left(\frac{5}{2^n+65}\right)= \left(\frac{3}{2^n+65}\right)\]from the definition of the jacobi symbol. Thus this is a contradiction.
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