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].