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Prove that $0_{1}+1_{2}+12_{3}+123_{4}+\cdots+(1:2:3:\cdots:2025)_{2026}$ is not a square, where ($1:2:3:\cdots:2025)_{2026}$ is the concatenation of all single digit in base $2026$.

My first instinct is to use $\pmod{10}$. I know that term $\pmod{10}$ repeats after $20$ terms, and the sum of last digit of first $20$ terms is $7$ (in decimal), so $(1:2:3:\cdots:2019)_{2020}$ is the $2020$th term in the sequence, and is the $101$st full cycle of the sequence. From that, I know that $(1:2:3:\cdots:2020)_{2021}$ must end with $0$ when converted to decimal.

But that's where I get stuck. I don't know how to proceed or if my approach is right.

Context: This question came in my mind after adding different numbers that are also different bases (example: $12_{4}+23_{7}$). Then I started to wonder that if we concatenate all single digits ($0$ is only included in base $1$, otherwise not) in base $b$ and add them, can we ever get a square number.

Note: This question did never came in any contest.

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  • $\begingroup$ FYI, the individual terms in the sum are given by oeis.org/A023811 $\endgroup$ Commented 14 hours ago
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    $\begingroup$ Base 10 is a poor choice for square detection, since 60% of the digits (0, 1, 4, 5, 6, and 9) may occur at the end of a square number. I recommend base 32 instead, because only 7/32 (21.875%) of the possible digits (0, 1, 4, 9, G (16), H (17), and P (25)) occur at the end of squares. $\endgroup$ Commented 14 hours ago
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    $\begingroup$ I don't know if this helps but $1234....(b-1)_b = 1111....1111_b + 1111....111_b + 1111...11_b+..... + 1=\frac 1{b-1}(b^{b-1}-1) + \frac 1{b-1}(b^{b-2} - 1)... +\frac 1{b-1}(b-1)=\frac 1{b-1}(11111....10_b)=\frac 1{b-1}(\frac {b^b-1}{b-1}-1)$. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ With computer assistance, I found that $S := \sum_{b=1}^{2026}\sum_{k=0}^{b-1} (b-1-k)b^k$ is divisible by the prime number $20011$, but not by $20011^2$. This means that $S$ is not a square number. I'm sure you want a less “brute-force” solution, though. $\endgroup$ Commented 14 hours ago
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    $\begingroup$ If I haven't made any typos or basic arithmetic errors, our sum is $\sum\limits_{b=2}^{2026}\frac 1{b-1}(\frac{b^b-1}{b-1} -b)$ or $\sum\limits_{k=1}^{2025}\frac 1k(\frac {(k+1)^{k+1}-1}k -(k+1))$. Not sure how to go from there. Induction might not be impossible. Do you think that any partial sum is ever a square. $\endgroup$ Commented 13 hours ago

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First let $$ T_k = (123...k)_{10} $$ $$ S =\sum_{k=1}^{2026} T_k $$ We must now prove that S is not a perfect square. We know that in base 10 by congruence $$ 10 \equiv 2 (mod 4), 10^2\equiv 0(mod4),10^m\equiv0(mod4)$$ For all m >=2. From this we can see when working in mod 4 we only depend the last two digits of our concatenated integer. $$ => T_k\equiv(\text Final \space two \space decimal \space digits \space of \space T_k) (mod4) $$ We can see for all k>= 10 the last two digits of T_k are the last two digits of k this gives us $$T_k\equiv k (mod4)$$ Now separate the sums $$S=\sum_{k=1}^{2026} T_k=\sum_{k=1}^{9} T_k+\sum_{k=10}^{2026} T_k $$ Now implement our mod 4 $$\sum_{k=10}^{2026} T_k \equiv \sum_{k=10}^{2026}k \equiv (\sum_{k=1}^{2026} k) - (\sum_{k=1}^{9} k)(mod4) $$ Compute first sum $$ \sum_{k=1}^{2026} k = 1013*2027 \equiv 1*3 \equiv 3(mod4) $$ From here we see 1013 is equivalent to 1(mod4) and 2027 3(mod4). We also see for k=45 we are still 1(mod4), so from here $$ \sum_{k=10}^{2026} T_k \equiv 3-1 \equiv 2(mod4) $$ Now for the first k=1..9 we simply observe the last two digits 1,12,23,34,45,56,67,78,89. Now we sum them and put the parts together $$ 1+12+23+34+45+56+67+78+89=405\equiv 1(mod4)$$ $$S\equiv \sum_{k=1}^{9} T_k + \sum_{k=10}^{2026} T_k \equiv 1+2+ \equiv 3(mod4) $$ Since we know a perfect square in mod4 can only be 1 or 0 and S is equivalent to 3(mod 4) we can see that S can not be a perfect square.

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  • $\begingroup$ Using MathJax formatting could help improve the answer; it may take time to learn but is worth the effort. For example $ k \ge 10 $ will display as $ k \ge 10 $; the line $$ => T_k\equiv(\text Final \space two \space decimal \space digits \space of \space T_k) (mod4) $$ would display better if rewritten as $$ \implies T_k \equiv \left(\text{final two decimal digits of } T_k\right) \pmod{4} $$ and that would display as $$ \implies T_k \equiv \left(\text{final two decimal digits of } T_k\right) \pmod{4} $$ $\endgroup$ Commented 6 hours ago

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