Existence of a "square-sum"-free word

Question

Let $A$ be a finite subset of an additive group $G$, with no pair of its elements being inverses. I would like to prove/disprove the existence of an infinite word over $A$ (treated now as an alphabet) that is free of "square-sums"--words of the form $CD$, where the sum of the letters of $C$ equals the sum of the letters of $D$ in $G$. Note that such a word would necessarily be square-free in the normal sense.

I've constructed some infinite square-free words using standard methods for some small values of $|A|$, and I suspect that such a word that is also square-sum-free cannot exist. Within these infinite words, I've been finding repeat sums for words of length $|A|$ or longer.

Example: The "Leech" square-free word

0121021201210120210201202120102101201021202102012021...

repeats 3 over and over via words of length 3, and I'm guessing it does so arbitrarily many times as the sequence progresses.

Addition:

I am especially interested in the situation where the sequence has a "tolerance" for repeat letters (but not repeat sums involving words consisting of more than one letter). That is, there exists at least one $a$ so that the words $aa, aaa$ are not considered squares/square-sums (but $aaaa$ is). Note that technically, such a sequence would no longer be square-free in the exact sense. For example, if $G=\mathbb{Z}$, then $6,6,6$ could be allowed but not $6,6,6,18$.

Answer 1

(Note: This answer was posted before the OP said that they are willing to "allow" words that contain subwords of the form $aa$ or $aaa$; i.e., that the "no consecutive subwords have the same sum" would require those consecutive subwords to have length $2$ or larger.)

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Too long for a comment. I suspect there is a cleverer way of doing this, but here is a way to show that $A$ would have to have at least four elements.

Say $A=\{a,b,c\}$ and we had such a word. Any set of four consecutive letters must use all three elements of $A$ (otherwise you will have either two consecutive letters the same, or a subword of the form $xyxy$, giving a square). In particular, whenever we have a pattern of the form $xyx$, the next letter must be the missing one. E.g., if we have $aba$, the next letter must be $c$.

I claim that our putative infinite word cannot have a pattern of the form $xyzx$ in four consecutive letters. Say, without loss of generality, that we have $abca$.

Case 1. Next letter is $b$, so we have $abcab$. Then the next letter cannot be $b$ or $c$ to avoid squares, so we must continue $abcaba$. Then the next letter must be a $c$ so the last four cover all three letters, but that gives $abcabac$, and then the second through fourth letters are the same as the fifth through seventh (one each). Thus, we end up with two consecutive sub-words that have the same totals for $a$, $b$, and $c$.

Case 2. The next letter is $c$, so we have $abcac$. The next letter must be a $b$ so that the last four cover all letters, so we get $abcacb$, but now we have the first three and last three letters are one each.

Thus, we cannot have $abca$ anywhere in the word, or more generally any four letters in the pattern $xyzx$.

So our initial pattern must be of the form $xyzy$ or $xyxz$.

In the latter case, say we have $abac$; then the next letter cannot be $b$ (that would give the pattern $xyzx$ in the last four), so we would have to have $abaca$. Then the last four letter have the pattern $xyzy$. So we can concentrate on that pattern.

Thus, say we have $abcb$. Then the next letter needs to be an $a$, to give $abcba$. If the next letter is $c$, we get the pattern $xyzx$ in the last four letters. So the next letter is $b$, to give $abcbab$. Then the next letter must be a $c$ so that the last four cover all three letters, giving $abcbabc$. The next letter cannot be $a$ (to avoid the $xyzx$ pattern), so it must be $b$, giving $abcbabcb$. Then looking at the $bcb$ end, we have that the next one must be $a$, giving $abcbabcba$; but this gives $bcba$ repeated at the end.

Thus, we must have $|A|\geq 4$.