PROBLEMS from volume 32 (1994) issue #4 of The Fibonacci Quarterly.

Copyright ◯^{c} 1994 by The Fibonacci Association.

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ELEMENTARY PROBLEMS AND SOLUTIONS Edited by Stanley Rabinowitz

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BASIC FORMULAS

The Fibonacci numbers F_{n} and the Lucas numbers L_{n} satisfy

F_{n+2} = F_{n+1} + F_{n}, F_{0} = 0, F_{1} = 1;
L_{n+2} = L_{n+1} + L_{n}, L_{0} = 2, L_{1} = 1.

Also, α = (1 + \sqrt{5})/2, β = (1 − \sqrt{5})/2, F_{n} = (α^{n} − β^{n})/\sqrt{5}, and L_{n} = α^{n} + β^{n}.

PROBLEMS PROPOSED IN THIS ISSUE

\underline{B–766} Proposed by R. André-Jeannin, Longwy, France

Let n be an even positive integer such that L_{n} ≡ 2 (mod p), where p is an odd prime. Prove that

L_{n+1} ≡ 1 (mod p).

\underline{B–767} Proposed by James L. Hein, Portland State University, Portland, OR Consider the following two mutual recurrences:

G_{1} = 1; G_{n} = F_{n+1}G_{n−1} + F_{n}H_{n−2}, n ≥ 2

and

H_{0} = 0; H_{n} = F_{n+1}G_{n} + F_{n}H_{n−1}, n ≥ 1.

Prove that H_{n−1} and G_{n} are consecutive Fibonacci numbers for all n ≥ 1.

\underline{B–768} Proposed by Juan Pla, Paris, France

Let u_{n}, v_{n}, and w_{n} be sequences defined by u_{1} = 1/2, v_{1} = \sqrt{2}, and w_{1} = (1/2)\sqrt{3}; u_{n+1} = u_{n}^{2} + v_{n}^{2} − w_{n}^{2}, v_{n+1} = 2u_{n}v_{n}, w_{n+1} = 2u_{n}w_{n}. Express u_{n}, v_{n}, and w_{n} in terms of Fibonacci and/or Lucas numbers.

\underline{B–769} Proposed by Piero Filipponi, Fond. U. Bordoni, Rome, Italy

Solve the recurrence

a_{n+1} = 5a_{n}^{3} − 3a_{n}, n ≥ 0

with initial condition a_{0} = 1.

\underline{B–770} Proposed by Andrew Cusumano, Great Neck, NY

Let U(x) denote the unit’s digit of x when written in base 10. Let H_{n} be any generalized Fibonacci sequence that satisfies the recurrence H_{n} = H_{n−1} + H_{n−2}. Prove that, for all n,

U(H_{n} + H_{n+4}) = U(H_{n+47}),

U(H_{n} + H_{n+5}) = U(H_{n+10}),

U(H_{n} + H_{n+7}) = U(H_{n+53}),

U(H_{n} + H_{n+8}) = U(H_{n+19}),

U(H_{n} + H_{n+11}) = U(H_{n+49}),

U(H_{n} + H_{n+13}) = U(H_{n+26}),

U(H_{n} + H_{n+16}) = U(H_{n+23}),

U(H_{n} + H_{n+17}) = U(H_{n+34}),

U(H_{n} + H_{n+19}) = U(H_{n+41}),

U(H_{n} + H_{n+20}) = U(H_{n+55}),

U(H_{n} + H_{n+23}) = U(H_{n+37}),

U(H_{n} + H_{n+25}) = U(H_{n+50}),

U(H_{n} + H_{n+28}) = U(H_{n+59}),

and U(H_{n} + H_{n+29}) = U(H_{n+58}).

\underline{B–771} Proposed by H.-J. Seiffert, Berlin, Germany Show that

∑_{n=1}^{∞} \frac{(2n + 1)F_{n}}{2^{n}n(n + 1)} = ln 4.