This is based on a classic problem.
Answer: his chain had from 8 to 15 links.
Suppose his chain had 7 links.
He could cut one link like this: .
Then he can present from 1 to 7 links.
Since he needed two cuts, he had more than 7 links,
. . up to a maximum of 15 links.
Suppose his chain had 15 links.
Again, he would cut the 3rd link . . . and cut the 7th link.
This gives him four chains with 2, 1, 4, and 8 links.
Then he can present from 1 to 15 links.