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$3\underbrace{88...8}_{n \text{ times}}1$ is $1$ mod $4$

$3\underbrace{88...8}_{n \text{ times}}1$ is sometimes $1$ mod $3$ (and sometimes divisible by 9)

$3\underbrace{88...8}_{n \text{ times}}1$ is sometimes $9$ mod $11$

I tried to find if are there any perfect squares that is form of $3\underbrace{88...8}_{n \text{ times}}1$, but I did not succeed.

Can $3\underbrace{88...8}_{n \text{ times}}1$ be a perfect square?

Thirdy Yabata
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    Have you tried anything? And what is the source of this problem? Is there any reason to consider this particular numerical form? Have you searched for any perfect powers of this form? – lulu Feb 04 '23 at 17:59
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    In any case, these are all $3\pmod 7$. Which, obviously, doesn't handle the perfect power case but is enough for squares. – lulu Feb 04 '23 at 18:07

2 Answers2

8

Reducing modulo seven, $381 \equiv 3$.

To get from this term to the next, note that $381(10) + 71 = 3881$.

More generally, new terms are generated by $n \mapsto 10n + 71$.

Again reducing modulo seven, observe that $3(10) + 71 = 101 \equiv 3$.

So, every term in your list is $3$ $\text{mod}$ $7$.

But, no square can have a remainder of three after division by seven.

Reducing modulo seven, we find:

$0^2 = 0 \equiv 0$

$1^2 = 1 \equiv 1$

$2^2 = 4 \equiv 4$

$3^2 = 9 \equiv 2$

$4^2 = 16 \equiv 2$

$5^2 = 25 \equiv 4$

$6^2 = 36 \equiv 1$.

Since three doesn't appear in this list, you cannot arrive at a square in your sequence.

Benjamin Dickman
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  • The implicit inductive proof above that $,N\equiv 3\pmod{7},$ can be viewed as a special case of the universal divisibility test, as I show in my answer. This simple idea often proves handy – Bill Dubuque Feb 04 '23 at 21:52
  • @BillDubuque Yes, I thought it wise to write an answer for this individual case but appreciate that you have a more general version. Unfortunately, we both seem to have suffered unnecessary downvotes! A shame that people are using MSE in this way... – Benjamin Dickman Feb 05 '23 at 18:32
  • Yes, alas, it appears that someone downoted both answers and the question within 10 seconds (that was the 2nd downvote on my answer). Likely that's because someone thinks the question is a PSQ. – Bill Dubuque Feb 06 '23 at 20:35
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Hint $ \bmod 7\!:\ \color{#0af}{38}\equiv \color{darkorange}3\ $ so $ \, \left\{\begin{align} \!N = &\ \overbrace{\color{#0af}{38}8\ldots81}^{\text{delete all } 8's}\\ \equiv &\ \ \ \color{darkorange}38\cdots81\\ &\ \ \ \quad\ddots \\ \equiv &\qquad\quad\ \color{#c00}{31}\end{align}\right\}\,$ so $\,N\equiv \color{#c00}{31}\equiv 3,\:\!$ which is $\rm\color{#c00}{non}$square $\!\bmod 7$

We applied the universal divisibility test, i.e. we reduce a natural $N\bmod 7\,$ by iteratively reducing its leading (two-) digit chunks, i.e. $\,\color{#0af}{38}\to \color{darkorange}3\, (= 38\bmod 7),\,$ which is a nice "absorbing" reduction that deletes any $\,8\,$ following the leading digit. Iterating, it eventually deletes all successive $8$ digits, leaving only $31$. This works in general to delete (absorb) all $\,b\,$ digits following the leading digit $\,a\,$ when $\bmod d\!:\ ab_{\phantom{}_{\large 10}}\!\equiv a,\,$ i.e. $\,10a+b\equiv a,\,$ i.e. $\,\color{#c00}{b\equiv -9a},\,$ as we prove inductively below.

For OP: $\ \bmod 7\!:\ \ \color{#c00}{8\equiv -9(3)},\ $ so $\ a_n = 3{88\ldots 88}_{\phantom{}_{\large 10}}\!\equiv N$ nonsquare, by $\,\color{#0af}{3\ \rm nonsquare}$, by:

Lemma $ \bmod d\!:\: $ if $\,\color{#c00}{b\equiv -9a}\,$ then $\,a_n = a{bb\ldots bb}_{\phantom{}_{\large 10}}$ is nonsquare $\iff \color{#0af}{a\ \rm is\ nonsquare}$

Proof $\ $ We induct on $\,n\,$ to show $\,\color{#0a0}{a_n\equiv a}\pmod{\!d}.\,$ Base case $\,n\!=\!0\,$ (no $b$'s) is true by $\,a_0 = a.\,$ Inductive step: $\ a_{k+1} = 10\color{#0a0}{a_k}+\color{#c00}{b}\equiv 10\color{#0a0}a-\color{#c00}{9a}\equiv a\,$ by $\rm\color{#0a0}{induction}$.

Bill Dubuque
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  • @(political?)Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate (even to political downvoters). It's truly unfortunate that nowadays many votes have more to do with site politics than with mathematics. – Bill Dubuque Feb 04 '23 at 19:36
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    Bill, Bill, people get random downvotes all the time. Don't be so sensitive! – TonyK Feb 04 '23 at 19:41
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    @TonyK In this case they are not random (a pattern of continuous (roughly) daily downvotes that started around a month ago). Alas, many users heavily involved in site moderation suffer such attacks from time to time. To the OP: the mathematics above is correct, and it shows how to generalize to many other repeated digit numbers. – Bill Dubuque Feb 04 '23 at 19:41
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    Note that "nonsquare" in the Lemma statement mean "nonsquare $!\bmod d,$ (i.e. the entire line is in the scope of the initial $!\bmod d,$ prefix). Possibly the downvoter misunderstood that and thought "nonsquare" meant "nonsquare in $\Bbb Z$". $\ \ $ – Bill Dubuque Feb 04 '23 at 19:54