# Fibonacci and Square Numbers - The Solution

Author(s):
Patrick Headley

Leonardo realizes that what he needs is a congruous number that is 5 times a square, because then he can divide through by that square and leave just the 5 [p. 76]. The values ab = 4 and bg = 5 almost work, since 4·5·(4+5)·(5-4) = 180 = 5·62. But 4+5 is not even as is required for a congruous number, the problem being that there are not 5 consecutive odd numbers centered at 36, or 4 centered at 45. Leonardo gets around this by doubling all the values. There are 10 consecutive odd numbers centered at 72, namely those from 63 to 81. There are also 8 consecutive odd numbers centered at 90, those from 83 to 97. Both sequences sum to 720=5·122, and we get

$41^2-720=(1+3+\ldots 81) - (63+65+ \ldots + 81) = 1 + 3 + \ldots + 61 = 31^2.$

and

$41^2+720=(1+3+\ldots 81) + (83+85+ \ldots + 97) = 49^2.$

Finally, Leonardo has his answer. He writes

There is for the first square 6 97/144, with root 2 7/12, which results from dividing 31 by the root of 144, which is 12, and there is for the second, which is the sought square, 11 97/144, with root 3 5/12, which results from dividing 41 by 12, and there is for the last square 16 97/144 with root 4 1/12 [p. 78].

In modern notation, we could write

$(41/12)^2-5=(31/12)^2.$

and

$(41/12)^2+5=(49/12)^2.$

Patrick Headley, "Fibonacci and Square Numbers - The Solution," Convergence (August 2011)