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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Nice inequality
TUAN2k8   2
N 10 minutes ago by TUAN2k8
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
2 replies
TUAN2k8
6 hours ago
TUAN2k8
10 minutes ago
An NT for a break
reni_wee   1
N 21 minutes ago by reni_wee
Source: ONTCP 2.4.1
Prove that there are no positive integers $x,k$ and $n \geq 2$ such that $x^2+1 = k(2^n -1)$.
1 reply
reni_wee
23 minutes ago
reni_wee
21 minutes ago
Inequality
lgx57   0
27 minutes ago
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
0 replies
lgx57
27 minutes ago
0 replies
p divides x^x-c
mistakesinsolutions   6
N 31 minutes ago by reni_wee
Show that for integer c and a prime p, $ p |x^x-c $ has a solution
6 replies
1 viewing
mistakesinsolutions
Jun 13, 2023
reni_wee
31 minutes ago
exponential diophantine in integers
skellyrah   1
N 33 minutes ago by skellyrah
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
1 reply
skellyrah
Yesterday at 7:04 PM
skellyrah
33 minutes ago
IMO 2017 Problem 4
Amir Hossein   117
N 37 minutes ago by ezpotd
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
117 replies
Amir Hossein
Jul 19, 2017
ezpotd
37 minutes ago
x^2+y^2+z^2+xy+yz+zx=6xyz diophantine
parmenides51   7
N an hour ago by Assassino9931
Source: Greece Junior Math Olympiad 2024 p4
Prove that there are infinite triples of positive integers $(x,y,z)$ such that
$$x^2+y^2+z^2+xy+yz+zx=6xyz.$$
7 replies
parmenides51
Mar 2, 2024
Assassino9931
an hour ago
Turkish JMO 2025?
bitrak   1
N 2 hours ago by blug
Let p and q be prime numbers. Prove that if pq(p+ 1)(q + 1)+ 1 is a perfect square, then pq + 1 is also a perfect square.
1 reply
bitrak
Yesterday at 2:04 PM
blug
2 hours ago
Combi Algorithm/PHP/..
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
5. [Czech and Slovak Republics 1997]
Each side and diagonal of a regular n-gon (n ≥ 3) is colored blue or green. A move consists of choosing a vertex and
switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the number of blue segments incident to each vertex even by following a sequence of moves. Also show that the final configuration obtained is uniquely determined by the initial coloring.
0 replies
CatalanThinker
2 hours ago
0 replies
Combi Proof Math Algorithm
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
3. [Russia 1961]
Real numbers are written in an $m \times n$ table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
0 replies
CatalanThinker
2 hours ago
0 replies
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   1
N 2 hours 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 viewing
Primeniyazidayi
3 hours ago
Ritwin
2 hours ago
Easy Taiwanese Geometry
USJL   14
N 3 hours 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
USJL
Jan 31, 2024
Want-to-study-in-NTU-MATH
3 hours ago
Problem 7
SlovEcience   6
N 3 hours 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
3 hours ago
Strange circles in an orthocenter config
VideoCake   1
N 3 hours 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
3 hours ago
a truly remarkable problem
john0512   15
N Apr 20, 2025 by cursed_tangent1434
Consider the sequence of positive integers $$6,69,696,6969,69696\cdots.$$It is well known that $69696=264^2$. Prove that this is the only perfect square in the sequence.

Jiahe Liu, Vikram Sarkar, Allen Wang, Ritwin Narra, Carlos Rodriguez, Susie Lu, Jonathan He, Jordan Lefkowitz, Victor Chen, Luv Udeshi
15 replies
john0512
Feb 20, 2023
cursed_tangent1434
Apr 20, 2025
a truly remarkable problem
G H J
G H BBookmark kLocked kLocked NReply
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john0512
4191 posts
#1 • 39 Y
Y by vsamc, CT17, awesomeguy856, iamhungry, GoodMorning, Jndd, fuzimiao2013, peelybonehead, purplepenguin2, ihatemath123, fidgetboss_4000, Arrowhead575, tauros, Lamboreghini, Taco12, asdf334, CyclicISLscelesTrapezoid, mahaler, Schur-Schwartz, megarnie, wenwenma, Mogmog8, alan9535, centslordm, Maximilian113, jasperE3, sky2025, Ritwin, EpicBird08, OronSH, Sedro, aidan0626, KnowingAnt, bjump, khina, vincentwant, Yiyj1, cursed_tangent1434, Ciobi_
Consider the sequence of positive integers $$6,69,696,6969,69696\cdots.$$It is well known that $69696=264^2$. Prove that this is the only perfect square in the sequence.

Jiahe Liu, Vikram Sarkar, Allen Wang, Ritwin Narra, Carlos Rodriguez, Susie Lu, Jonathan He, Jordan Lefkowitz, Victor Chen, Luv Udeshi
This post has been edited 2 times. Last edited by john0512, Feb 20, 2023, 4:14 PM
Reason: ok taco12
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HumanCalculator9
6231 posts
#2
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It is well-known that no number consisting of 69 concatenated with itself n times is a square (see Conway paper)

Now time to prove that 6969...6 cases
insert fakesolve here
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Math2019A
254 posts
#3
Y by
outline?
actually nvm this doesn't work
This post has been edited 1 time. Last edited by Math2019A, Feb 20, 2023, 3:32 AM
Reason: ig
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peelybonehead
6290 posts
#4
Y by
Insane problem now prove by fake induction
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ihatemath123
3449 posts
#6 • 2 Y
Y by OronSH, Yiyj1
When one solves this problem, they both feel satisfaction at having acheived the goal, but also sorrow for the fact that there is no hope at discovering a second funny number...
This post has been edited 3 times. Last edited by ihatemath123, Feb 20, 2023, 4:35 AM
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BorealBear
399 posts
#8
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HumanCalculator9 wrote:
It is well-known that no number consisting of 69 concatenated with itself n times is a square (see Conway paper)

Now time to prove that 6969...6 cases
insert fakesolve here

The $ 6969...96 $ case is actually trivial. Consider $ n=6969...96 $ and $ n>69696 $. Then, notice that $ 2^5|n $ but $ 2^6 $ does not divide $ n $. Thus, it is impossible for $ n $ to be a perfect square. Then test the ones not eliminated.
This post has been edited 1 time. Last edited by BorealBear, Oct 1, 2023, 3:48 AM
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firebolt360
903 posts
#9
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Woah wait this problem is solved...
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IAmTheHazard
5003 posts
#11 • 20 Y
Y by Jndd, Lamboreghini, Schur-Schwartz, john0512, samrocksnature, purplepenguin2, CyclicISLscelesTrapezoid, GrantStar, ihatemath123, mahaler, channing421, CoolCarsOnTheRun, Mogmog8, centslordm, EpicBird08, Sedro, aidan0626, khina, OronSH, Yiyj1
We seek the positive integer solutions $n$ to either
$$\frac{69\cdot 10^{2k+1}-100+4}{99}=n^2$$or
$$\frac{69\cdot 10^{2k}-100+31}{99}=n^2.$$The first equation rearranges to
$$230\cdot 100^k=33n^2+32.$$It is clear that $\nu_2(33n^2+32) \in\{0,2,4,5\}$, so $1+2k \leq 5 \implies k \leq 2$. Checking these gives the solution $n=264$ only, which is what we expected.
The second equation rearranges to
$$23\cdot 100^k=33n^2+23.$$Evidently $23 \mid n^2$, so let $n=23m$, so
$$100^k-1=759m^2 \implies (10^k)^2-759m^2=1.$$Solutions $(k,m)$ to this equation are solutions $(x,y)=(10^k,m)$ to the Pell equation $x^2-759y^2=1$, which has (positive) solutions
$$(x,y)=\left(\frac{(551+20\sqrt{759})^a+(551-20\sqrt{759})^a}{2},\frac{(551+20\sqrt{759})^a-(551-20\sqrt{759})^a}{2\sqrt{759}}\right),$$where $a$ is a nonnegative integer. On the other hand, we have
$$\frac{(551+20\sqrt{759})^a+(551-20\sqrt{759})^a}{2}=\sum_{i=0}^{\lfloor a/2 \rfloor} \binom{a}{2i}303600^{2i}\cdot 551^{a-2i} \equiv \binom{a}{0}551^a \equiv 1 \pmod{10},$$so any solution $(x,y)$ to the aforementioned Pell equation cannot possibly have $x$ be a power of $10$ (aside from $1$, which we exclude since it implies $m=0$). Thus $69696=264^2$ is the only square in the sequence. $\blacksquare$
This post has been edited 3 times. Last edited by IAmTheHazard, Feb 21, 2023, 3:32 PM
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john0512
4191 posts
#12
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non-pell

also i feel like this should be in HSO lol
This post has been edited 1 time. Last edited by john0512, Feb 21, 2023, 10:11 PM
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thdnder
198 posts
#13
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Nice application of LTE lemma.

Assume the contrary. Let $n$ be the number of digits. We divide in 2 cases:

Case 1: $n$ is odd.

Let $n = 2k + 1$. Then $6969\dots 696 = 69(\frac{10^{2k} - 1}{10^2 - 1}) \cdot 10 + 6 \equiv 69 \frac{-1}{99} \cdot 10 + 6 \equiv \frac{-32}{33} (2^{2k})$. Thus if $n \ge 7$, then $\nu_{2}(6969\dots 696) = 5$, a contradiction. $n = 1, 3, 5$ cases are checked by hand.

Case 2: $n$ is even.

Then our number becomes $69\frac{10^{n} - 1}{10^2 - 1}$. Since $23 \mid \frac{10^{n} - 1}{10^2 - 1}$, we have $11 \mid n$. It's not hard to check that if $s > 1$, then $10^s - 1$ is not perfect square. (If $s$ is even, it's obvious, else $10^s - 1$ is slightly greater than perfect square) Define sequence $(p_{n})_{n \ge 0}$ such that $p_{0} = 11$ and for $i \ge 1$, $p_{i}$ is prime factor of $10^{p_{i - 1}} - 1$ such that $\nu_{p_{i}}(10^{p_{i - 1}} - 1)$ is odd. $p_{1} \mid 10^{11} - 1 \mid \frac{10^n - 1}{10^2 - 1}$, so by lifting the exponent, we have $\nu_{p_1(n)}$ is odd, so $p_{1} \mid n$. A induction on $i$ gives $p_{i} \mid n$. Thus $n$ is divisible by $p_{1}p_{2}\dots$, a contradiction. $\blacksquare$
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popop614
272 posts
#16
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After two fakesolves, I swear this is right.

Handle all the $69$s first.
\[ \frac{23(100^n - 1)}{33} = m^2. \]
Assume $n$ is odd. Rule out $n=1$. Then we can split as follows:
\[ 23\left( \frac{10^n - 1}{3} \right) \left( \frac{10^n + 1}{11} \right) = m^2. \]Suppose
\[ 230 \cdot 10^{2k} + 23 = 11m^2. \]Then $23 \mid m$ so let $m = 23k$. As such,

\[ 10 \cdot 10^{2k} + 1 = 253m^2. \]Take mod $8$. On the LHS we get $1$ or $3$. The RHS on the other hand gives us $5m^2$. This is only ever gives us $0$, $4$, or $5$ mod $8$. Bad.

Now suppose
\[ 10 \cdot 10^{2k} - 1 = 3m^2. \]Same argument suffices. On the LHS we either get $1$ or $7$ mod $8$, the RHS we only ever get $0$, $3$, or $4$. This never happens.

We have two relatively prime (remark that $23 \nmid 10^n - 1$) non-perfect squares multiplying to a perfect square. This is impossible.

If $n$ is even it becomes

\[ 23 \left( \frac{10^n - 1}{33} \right) \left( 10^n + 1 \right). \]Collating the $23$ with the first one yields the desired result, since we get relatively prime factors which are $3$ mod $4$.

For the other cases, verify that
\[ \nu_2(6969696) = 5 \]and then
\[ \nu_2(69 \cdot 10^{7}) > 5\]and so it suffices to check $6$, $696$, and $69696$. We easily discard the first two, and the latter is $264^2$ as given.
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Mathandski
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#17
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Funny number problem for 400th post.
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shendrew7
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#18
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First consider an even $2k$ number of digits. Then our number can be expressed as
\[69(10^{2k-2}+10^{2k-4}+\ldots+1) = \frac{23}{3 \cdot 11}(10^k-1)(10^k+1).\]
Notice that $\gcd(10^k-1,10^k+1)=1$, so each of the primes 3, 11, and 23 divide exactly one:
  • $k$ even: Then $11 \mid 10^k-1 \implies 23 \mid 10^k+1 \implies k \equiv 11 \pmod{22}$, contradiction.
  • $k$ odd: Then $11 \mid 10^k+1 \implies 23 \mid 10^k-1 \implies k \equiv 0 \pmod{22}$, contradiction.

When we have an odd number of digits, notice $v_2(69696)=6$ and for all succeeding terms,
\[v_2(69\ldots69696) = v_2(69\ldots696 \cdot 100 + 96) = v_2(96) = 5 \not\equiv 0 \pmod 2. \quad \blacksquare\]
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bjump
1035 posts
#19
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I have been enlightened by this truly remarkable problem.

Suppose we have $2n+2$ for $n \ge 0$ digits then our number can be written in the form: $$\frac{23(10^{2n}-1)}{33} = \frac{23(10^n-1)(10^n+1)}{33}$$Observe that $\gcd(10^n-1, 10^n+1) =1$. So we can split into cases.

Case 1: $10^n +1 = 23a^2$, and $10^n-1 = 33b^2$ we can easily check that $n=0,1,2$ give us no squares. So $23a^2 \equiv 1 \pmod{8} \implies a^2 \equiv 7 \pmod{8}$ which is an NQR mod $8$ contradiction.

Case 2: $10^n + 1 = 253 a^2$, $10^n -1 = 3b^2$. We have $ 3b^2 \equiv -1 \pmod{8} \implies b^2 \equiv 5 \pmod{8}$ which is a NQR mod $8$ contradiction.

Case 3: $10^n +1 = 11a^2$, $10^n -1 = 69 b^2$. Which fails because $1 \equiv 3a^2 \pmod{8} \implies a^2 \equiv 7 \pmod{8}$ a contradiction because $7$ is an NQR mod $8$.

Case 4: $10^n+1 = a^2$, $10^{n}-1 = 3 \cdot 11 \cdot 23 b^2$. Modulo $11$ gives $n$ even so substitute $n=2n_1$, we get the same equation we are trying to solve. Since cases $1-3$ fail we have $n_1$ even. Repeating this argument this case fails by infinite descent .

Now Suppose we have $2n+1$ digits
Observe that $6$, and $696$ fails. $69696 = 264^2$ and has a $\nu_2$ of $6$, also note that $9696$ has a $\nu_2$ of $5$. However if $n\ge 3$ then $\nu_2(69...69696) = \nu_2 ( 69...690000 + 9696)=5$ contradiction.
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OronSH
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#20
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First note no square ends in $969696$, as otherwise it is $32\pmod{64}$. Additionally check that $6,696$ fail.

Next write $a^2=\frac{23}{33}(b^2-1)$. We claim $b$ is not a power of $10$.

Rearrange this to $23b^2-33a^2=23$. It follows that $23\mid a$, so write $a=23c$ to get $b^2-759c^2=1$. This is a Pell equation, and we can see the smallest solution is $(b,c)=(551,20)$. Thus the general solution is given by $b+c\sqrt{759}=(551+20\sqrt{759})^n$. However now we see $b\equiv 1\pmod{10}$, failing.

This rules out all cases, so we are done.
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cursed_tangent1434
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#21
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Let $(a_n)$ denote the given sequence. We now split into two cases based on the parity of $n$.

Case 1 : $n$ is odd. Here note that the term ends with the digit 6 and hence is always even. We look at the following key claim.

Claim : For all positive integers $i$, the number \[N_i=9696\dots96 \ \ \ \ (96 \text{ repeated } i \text{ times}) \]satisfies $\nu_2(N_i)=5$.

Proof :
We approach this via induction. Clearly $96 = 3\cdot 32$ so $\nu_2(96)=5$. Now, say we have shown that $\nu_2(N_k)=5$ for some $k \ge 1$. Note,
\[N_{k+1}=96\cdot 10^{2k}+N_k\]so
\[\nu_2(N_{k+1}) = \min\{96\cdot 10^{2k},N_k\}=5\]since $64 \mid 96\cdot 10^{2k}$ for all $k \ge 1$. This completes the induction and the claim is proved.

However note that for all $i\ge 3$,
\[a_{2i+1} = 6\cdot 10^{2i}+N_{i}\]so
\[\nu_2(a_{2i+1})=\min\{6\cdot 10^{2i},N_i\}=\nu_2(N_i)=5\]since $\nu_2(6\cdot 10^{2i}) = 2i+1>5$ for all $i \ge 3$. Thus, no odd $n \ge 7$ may be a square. It is not hard to check that $a_1=6$ and $a_3=696$ are not squares, so the only square when $n$ is odd is $a_5=69696=264^2$.

Case 2 : $n$ is even. Here note that the term is a concatenation of the block $69$ so we may write,
\[a_{2i} = 6969\dots 69 = 69\cdot 10^{2i-2}+69\cdot 10^{2i-4}+\dots + 69= 69 \left(\frac{100^{i}-1}{100-1}\right) = \frac{23}{33} \left(10^{2i}-1\right)\]And hence the equation we wish to resolve is,
\[\frac{23}{33}\left(10^{2i}-1\right) = x^2\]for some positive integer $x$. For this note that,
\[x^2 = \frac{23}{33}\left(10^{2i}-1\right) = \frac{23}{33}(10^i-1)(10^i+1)\]where the two bracketed terms are successive odd integers and are hence relatively prime. But this means, we can split into four cases.

If $23 \cdot 33 \mid 10^i-1$ then we must have $10^i+1=t^2$ for some positive integer $t$. But by Mihaelescu's Theorem this has no solutions.

If $\gcd(10^i-1,23\cdot 33)=1$ then we must have $10^i-1=t^2$ for some positive integer $t$. Once again this has no solutions for $i>1$ by Mihaelescu's Theorem and when $i=1$ , $a_2=69$ is most clearly not a perfect square.

If $\gcd(10^i-1,23\cdot 33)=23$ then we must have $10^i+1=33\cdot t^2$ for some positive integer $t$. But this means $t^2 \equiv 7 \pmod{10}$ which is clearly not a quadratic residue $\pmod{10}$. Similarly if $\gcd(10^i-1,23\cdot 33)=33$ then we must have $10^i+1=23\cdot t^2$ for some positive integer $t$ which has no solutions for the exact same reasons.

Thus, in all cases we have no valid solutions and we are done.
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