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#1 h Feb 20, 2023, 3:21 AM • 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

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HumanCalculator9

#2 h Feb 20, 2023, 3:25 AM

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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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254 posts

#3 h Feb 20, 2023, 3:31 AM

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outline?
actually nvm this doesn't work

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Reason: ig

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6291 posts

#4 h Feb 20, 2023, 3:42 AM

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Insane problem now prove by fake induction

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3446 posts

#6 h Feb 20, 2023, 4:06 AM • 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...

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#8 h Feb 20, 2023, 5:15 AM

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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.

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#9 h Feb 20, 2023, 5:39 AM

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Woah wait this problem is solved...

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5001 posts

#11 h Feb 20, 2023, 1:15 PM • 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. $\blacksquare$

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#12 h Feb 21, 2023, 10:09 PM

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non-pell

also i feel like this should be in HSO lol

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#13 h Sep 29, 2023, 4:43 PM

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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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#16 h Nov 23, 2023, 3:58 AM

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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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#17 h Aug 26, 2024, 8:02 PM

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Funny number problem for 400th post.
Subjective Rating (MOHs) $\text{[math]}$

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#18 h Sep 9, 2024, 2:55 AM

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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

#19 h Dec 27, 2024, 4:31 AM

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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

#20 h Mar 27, 2025, 10:16 PM

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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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cursed_tangent1434

#21 h Apr 20, 2025, 12:26 PM

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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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