The terms of a sequence of positive integers satisfy a_{n+3} = a_{n+2}(a_{n+1} + a_{n}), for n = 1, 2, 3, ... .
If a_{6} = 8820, what is a_{7}?
Letting a_{1} = x, a_{2} = y, and a_{3} = z, from the recurrence relation we obtain
a_{4} = z(y + x),
a_{5} = z(y + x)(z + y),^{ }
a_{6} = z(y + x)(z + y)[z(y + x) + z] = z^{2}(y + x)(y + x + 1)(z + y).
Hence we have z^{2}(y + x)(y + x + 1)(z + y) = 8820 = 2^{2} × 3^{2} × 5 × 7^{2}. A certain amount of trial and error is now required. The goal is to minimize the need for trial and error by applying mathematical constraints.
By the Fundamental Theorem of Arithmetic, the factors on the lefthand side are the same as those on the righthand side of the equation. In particular, two of the factors on the lefthand side are consecutive integers, and therefore must be relatively prime. (Any divisor of n and n + 1 must divide n + 1 − n = 1.) This enables us to determine candidate values for (y + x) and (y + x + 1)  by partitioning the prime factors 2, 3, 5, 7  more easily than if we had to obtain a complete list of the 54 factors of 8820.
Additionally, we note that if z = 1, then (y + 1)(y + x)(y + x + 1) = 8820, with y + 1 y + x < y + x + 1, so that we must have y + x + 1 > 8820^{1/3}; that is, y + x 20. This enables us to exclude the case z = 1 from many of the candidate factorizations below.
(y + x)  (y + x + 1)  z^{2}(z + y)  Deductions 

2^{ }  3^{ }  2 × 3 × 5 × 7^{2}  y + x = 2 y = 1. 1 < z^{2}  2 × 3 × 5 × 7^{2} z = 7. Hence z^{2}(z + y) = 7^{2} × 8. Contradiction. 
3^{ }  2^{2}  3 × 5 × 7^{2}  y + x = 3 y 2. 1 < z^{2}  2 × 3 × 5 × 7^{2} z = 7. Hence z^{2}(z + y) 7^{2} × 8. Contradiction. 
2^{2}  5^{ }  3^{2} × 7^{2}  y + x = 4 y 3. 1 < z^{2}  3^{2} × 7^{2} z = 3, 7, or 21. Then:

5^{ }  2 × 3^{ }  2 × 3 × 7^{2}  1 < z^{2}  2 × 3 × 7^{2} z = 7. Hence z^{2}(z + y) > 7^{3}. Contradiction. 
2 × 3  7  2 × 3 × 5 × 7  1 < z^{2}  2 × 3 × 5 × 7 is impossible. 
3^{2}  2 × 5^{ }  2 × 7^{2}  1 < z^{2}  2 × 7^{2} z = 7. Hence z^{2}(z + y) > 7^{3}. Contradiction. 
2 × 7  3 × 5  2 × 3 × 7  1 < z^{2}  2 × 3 × 7 is impossible. 
2^{2} × 5  3 × 7^{ }  3 × 7^{ }  1 < z^{2}  3 × 7 is impossible. 
5 × 7^{ }  2^{2} × 3^{2}  7^{ }  z^{2}  7 z = 1, and y = 6, x = 29. Solution! 
Hence we have two solutions, (x, y, z) = (2, 2, 7) or (29, 6, 1). The two sequences are thus, respectively:
2, 2, 7, 28, 252, 8820, 2469600... , and
29, 6, 1, 35, 245, 8820, 2469600... .
Therefore, if a_{6} = 8820, a_{7} = 2469600.
Curiously, although we are able to deduce the value of a_{7}, we cannot uniquely determine a_{8}.
Source: Adapted from More Mathematical Morsels (Olympiad Corner, 1986), by Ross Honsberger