Are there any perfect squares that is form of $3\underbrace{88...8}_{n \text{ times}}1$?

Question

$3\underbrace{88...8}_{n \text{ times}}1$ 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?

Answer 1

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.

Answer 2

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 $\rm\color{#0a0}{induction}$.