Problem 11968
(American Mathematical Monthly, Vol.124, March 2017)
Proposed by C. J. Hillar (USA), R. Krone (Canada), and A. Leykin (USA).
For n ≥ 1, prove that F5n/(5Fn) is an integer congruent to 1 modulo 10 where Fn is the nth Fibonacci number.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. It suffices to show that F5n
5Fn
= 5Fn2(Fn2+ (−1)n) + 1.
In fact, the right-hand side is an integer and
5Fn2(Fn2+ (−1)n) + 1 ≡ 5Fn2(Fn2+ 1) + 1 ≡ 1 (mod 10) because the product of two consecutive integers Fn2(Fn2+ 1) is even.
Let ϕ± = (1 ±√
5)/2, x = ϕn+ and y = ϕn− then xy = (−1)n and Fn = (x − y)/√ 5.
Moreover
5Fn2= (x − y)2= x2− 2xy + y2= x2+ y2− 2(−1)n⇒ x2+ y2= 5Fn2+ 2(−1)n, and
25Fn4= (x − y)4= x4− 4x3y + 6x2y2− 4xy3+ y4
= x4+ y4− 4(−1)n(x2+ y2) + 6 ⇒ x4+ y4= 25Fn4+ 4(−1)n(x2+ y2) − 6.
Therefore F5n
Fn
= x5− y5
x − y = x4+ x3y + x2y2+ xy3+ y4= (x4+ y4) + (−1)n(x2+ y2) + 1
= 25Fn4+ 5(−1)n(x2+ y2) − 5 = 25Fn4+ 5(−1)n(5Fn2+ 2(−1)n) − 5
= 25Fn4+ 25(−1)nFn2+ 5 = 55Fn2(Fn2+ (−1)n) + 1
and the proof of the required identity is complete.
