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A pen and paper challenge:
Write down a number whose square (repetition allowed) is the concatenation of at least 25 nonzero squares.

examples
7, 13, 12: Their squares are the concatenation of 2 squares. (4,9 and 16,9 and 1,4,4 resp.)
10: Its square is the concatenation of 3 squares (1,0,0) but only 1 nonzero square (i.e. no concatenation at all) for the purpose of this question! The two 0s do not count.

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2 Answers 2

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Here is a simple solution

40204000002010200000402042 = 16163616161616361616363681363616163616161616361616

which is the concatenation of

81, 8 x 36 and 16 x 16

and entirely built on

2122 = 44944

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  • $\begingroup$ Could you explain how you 'found'/why you chose 212. (12 seems a more obvious choice - and is scales - and also using 12 allows even a smaller number) I did pick 25 esp. for this answer though, with only 2 digit squares it is very nice i.m.o. $\endgroup$
    – Retudin
    Sep 23, 2022 at 16:11
  • $\begingroup$ @Retudin Using 1s and 2s makes sense because of the prefactor 2 in the binomial expansion. Avoiding deep nesting of the Kronecker product is desirable because it minimizes the complications arising from different length squares in the concatenation. 212^2 consisting of only square digits that are not too different in size is just lucky as far as I can tell. (The four 4s are there by design but the 9 is coincidence). $\endgroup$
    – loopy walt
    Sep 23, 2022 at 20:42
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Consider the sequence $a_n = \prod_{k=0}^{n-1}(10^{3^k}+2)$. The sequence begins like this:
$$a_0 = 1, a_1 = 12, a_2 = 12024, a_3 = 12024000024048, \ldots$$

By induction, we observe the following: $a_n^2$ has exactly $3^n$ digits, the first of which is $1$, so $4a_n^2$ also has exactly $3^n$ digits.

Since $a_n^2 = \prod_{k=0}^{n-1}(100^{3^k}+4\cdot 10^{3^k}+4) = (100^{3^{n-1}}+4\cdot 10^{3^{n-1}}+4)a_{n-1}^2$, this implies that $a_n^2$ is formed by concatenating $a_{n-1}^2$, $4a_{n-1}^2$, and $4a_{n-1}^2$ without zeros in between. This means that $a_n^2$ is always the concatenation of two more squares than $a_{n-1}^2$, and since $a_0^2 = 1$ is the concatenation of one square, the number $a_{12}^2$ is the concatenation of $25$ squares.

Solution:

$a_{12} = \prod_{k=0}^{11}(10^{3^k}+2)$ satisfies the requirements. Its square begins as follows:
$a_{12}^2 = \color{red}{1}4\color{red}{4}576\color{red}{576}578306304\color{red}{578306304}578306306313225218313225216\ldots$


Better solution:

Consider the sequence $b_n = 12\cdot \prod_{k=0}^{n-1}(10^{6 \cdot 5^k}+10^{3 \cdot 5^k}/2 + 1)$. With the same reasoning as before, you see that $b_n$ has $3 \cdot 5^n$ digits, but now each $b_n^2$ is a concatenation of $b_{n-1}^2, b_{n-1}^2, \frac 94 b_{n-1}^2, b_{n-1}^2, b_{n-1}^2$, so it is made of four times as many squares as $b_{n-1}^2$, plus one. So if $b_0^2$ is made of three squares ($1, 4, 4$), $b_1^2$ is made of $13$ squares and $b_2^2$ is made of $53$ squares.

Solution: $b_2 = 12006012000000006003006000000012006012$ and $b_2^2 = 144144\color{red}{324}144144144144\color{red}{324}144144\color{red}{324324729324324}$ $144144\color{red}{324}144144144144\color{red}{324}144144$. Observe that we can coincidentally separate $324324729324324$ into five squares so we can actually separate this square into $57$ squares.

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  • $\begingroup$ Nice constructive approach, but your final answer looks like it is actually a(12)^2. I think you're missing a "^2" in a few places $\endgroup$
    – fljx
    Sep 22, 2022 at 12:50
  • $\begingroup$ This is indeed a nice constructive solution; similar to what I had in mind. But the challenge is to write down the number, not a formula for the number, and I dare you to actually write down the 3^12 digits. Find a better sequence and writing down the answer should not be a problem. $\endgroup$
    – Retudin
    Sep 22, 2022 at 14:42
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    $\begingroup$ I have found a better sequence. $\endgroup$
    – Magma
    Sep 22, 2022 at 16:11

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