Conjecture: For every word of length n, the maximum number of distinct squares that occur as factors of it is bounded by n.
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 , 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  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  the authors show that there is an exponential dependency between the number of squares and the number of holes a word has, while in , 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 ).