*Proof:* Let the lengths of the *3* squares be *2a*, *2b*, and *2c* with a < b < c. Let *s(x)* denote the symbol in position *x* in the string, with position *0* being the leftmost. Thus:

* s(x) = s(x+a) * for * 0 ≤ x < a (0) *

* s(x) = s(x+b) * for * 0 ≤ x < b (1) *

* s(x) = s(x+c) * for * 0 ≤ x < c (2) *

If *c ≥ 2a*, then *(2)* implies that *s(x) = s(x+c)* for *0 ≤ x < 2a*, so the square of length *2a* also occurs in positions *c* to *c + 2a - 1*, and the proof is complete in this case.

For the rest of the proof, assume that *c < 2a*. This implies that

* 0 < c - b < 2a - b < a < b < c < 2a < a + b*

I claim that * s(x) = s(x+b-a)* for *0 ≤ x < 2a (3)*.

Assuming this, the square of length *2a* also occurs in positions *b-a* to *a+b-1*, completing the proof.

To prove *(3)*, we consider 3 cases:

*A)* If *0 ≤ x < 2a - b*. Then *s(x) = s(x+b)* since * 0 ≤ x < b*
equals *s(x+b-a)* since *0 ≤ x + b - a < a*, and hence *s(x) = s(x+b-a)*.

*B)* If *a ≤ x < 2a*. Then *s(x) = s(x-a)* since *0 ≤ x-a < a*, thus *s(x)= s(x+b-a)* since *0 ≤ x-a < b*, and hence *s(x) = s(x+b-a)*.

*C)* If *2a - b ≤ x < a*, we proceed by induction on *x*. The base case relies on *A)*.

Assume that *s(y) = s(y+b-a)* for all *y* such that *0 ≤ y < x (4)*

Then,
*s(x) = s(x+b)* since *0 ≤ x < b*, thus *s(x) = s(x+b-c)* since *0 ≤ x + b - c < c*, thus *s(x) = s(x+2b-a-c)* by *(4)* with *0 ≤ y = x + b - c < x*, thus *s(x) = s(x+2b-a)* since *0 ≤ x + 2b - a - c < c*, thus *s(x) = s(x+b-a)* since *0 ≤ x + b - a < b*, so *s(x) = s(x+b-a)*, and the induction is complete.

The three cases cover the range *0 ≤ x < 2a*, so *(3)* is proved, and therefore we can conclude. ◼

The problem was further investigated in **[4-7]** with all papers presenting different possible strategies or techniques. In **[5]**, the authors consider different types of squares by looking at the number of squares starting at the same position. They call two squares starting at the same position an FS-double-square, and bound the number of these to *5n/6*. As a result, they manage to obtain the best known bound so far (*11n/6*) for the number of distinct squares that are factors of a word.
Another important breakthrough came in 2015 when Manea and Seki show in **[8]** that the conjectured fact that the maximum ratio between the length of the word and the number of distinct squares that it contains as factors is actually obtained on a binary alphabet.

As a final remark, note that the problem has been further studied also in the context of partial words. Thus in **[9]** the authors show that there is an exponential dependency between the number of squares and the number of holes a word has, while in **[10]**, the authors prove that when you consider binary words with only one hole, the bound is *7n/2* (this result was also proved independently in **[11]**).

This result was **settled** by Srecko Brlek and Shuo Li in a paper in 2022 **[12]**. Li has continued the investigation in a series of papers, and has further improved on the initial conjecture, provind also a conjecture of Manea and Seki **[13]**.

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- Srečko Brlek, Shuo Li: On the number of squares in a finite word. https://doi.org/10.48550/arXiv.2204.10204
- Shuo Li: n upper bound of the number of distinct powers in binary words. https://doi.org/10.48550/arXiv.2209.06891