How can we find two squares that sum to a square? Leonardo answers this question in several ways, with the first method following from a simple observation that provides inspiration throughout the book:

I thought about the origin of all square numbers and discovered that they arise out of the increasing sequence of odd numbers; for the unity is a square and from it is made the first square, namely 1; to this unity is added 3, making the second square, namely 4, with root 2; if to the sum is added the third odd number, namely 5, the third square is created, namely 9, with root 3; and thus sums of consecutive odd numbers and a sequence of squares arise together in order [p. 4].

Thus, the sums 1 + 3 + 5 + 7 = 16 and 1 + 3 + 5 + 7 + 9 = 25 are both squares. Since we add the square 9 to the first sum in order to get the second, we have 16 + 9 = 25 as a sum of two squares adding to a third square. Leonardo explains that we could use any odd square in place of 9 to do the same thing. For instance, using 49, we have 1 + 3 + ... + 47 = 576 and 1 + 3 + ... + 49 = 625, so 576 + 49 = 625 is another sum of the same form. Leonardo goes on to note that the final two or more terms of these odd-number sums can also sum to a square. For instance, an even square can be partitioned into consecutive odd numbers, an example being 17 + 19 = 36. Since 1 + 3 + ... + 15 = 64 and 1 + 3 + ... + 19 = 100, we get 64 + 36 = 100.

Taking things further, Leonardo poses the following problem:

I wish to find three squares so that the sum of the first and the second as well as all three numbers are square numbers [p. 105].

He explains his solution this way:

I shall find first two square numbers which have sum a square number and which are relatively prime. Let there be given 9 and 16, which have sum 25, a square number. I shall take the square which is the sum of all odd numbers which are less than 25, namely the square 144, for which the root is the mean between the extremes of the same odd numbers, namely 1 and 23. From the sum of 144 and 25 results, in fact, 169, which is a square number. And thus is found three square numbers for which the sums of the first two and all three together are square numbers [p. 105].

In fact, Leonardo points out that this method can be extended to any number of squares, since (1 + 3 + ... + 167) + 169 = 7056 + 169 = 84^{2} + 13^{2} = 85^{2} = 7225, and (1 + 3 + ... + 7223) + 7225 = 3612^{2} + 85^{2} = 3613^{2}. Thus, we get the following sequence of squares:

\[\begin{array}{ccc}3^2&=&3^2,\\3^2+4^2&=&5^2,\\3^2+4^2+12^2&=&13^2,\\3^2+4^2+12^2+84^2&=&85^2,\\3^2+4^2+12^2+84^2+3612^2&=&3613^2,\end{array}\]

and so on.