Square Fibonacci Numbers, Etc.

JOHN H. E. COHN

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This paper originally appeared as:

J. H. E. Cohn, "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 1964, pp. 109-113.

Square Fibonacci Numbers, Etc.

JOHN H. E. COHN

Bedford College, University of London, London, N.W.1.

INTRODUCTION

An old conjecture about Fibonacci numbers is that 0, 1 and 144 are the only perfect squares. Recently there appeared a report that computation had revealed that among the first million numbers in the sequence there are no further squares [1]. This is not surprising, as I have managed to prove the truth of the conjecture, and this short note is written by invitation of the editors to report my proof. The original proof will appear shortly in [2] and the reader is referred there for details. However, the proof given there is fairly long, and although the same method gives similar results for the Lucas numbers, I have recently discovered a rather neater method, which starts with the Lucas numbers, and it is of this method that an account appears below. It is hoped that the full proof together with its consequences for Diophantine equations will appear later this year. I might add that the same method seems to work for more general sequences of integers, thus enabling equations like
y2 = D x4 + 1
to be completely solved for certain values of D. Of course the Fibonacci case is simply D = 5.

PRELIMINARIES

In the first place, in accordance with the practice of the Fibonacci Quarterly, I here use the symbols Fn and Ln to denote the n-th Fibonacci and Lucas number respectively; in other papers I use the more widely accepted, if less logical, notation un and vn [3]. Throughout the following, n, m, k will denote integers, not necessarily positive, and r will denote a non-negative integer. Also, wherever it occurs, k will denote an even integer, not divisible by 3. We shall then require the following formulae, all of which are elementary

(1)2 Fm+n= Fm Ln + Fn Lm
(2)2 Lm+n= 5 Fm Fn + Lm Ln
(3) L2m= Lm2 + (-1)m-1 2
(4)(F3m, L3m) = 2
(5)(Fn, Ln) = 1   if 3 does not divide n
(6) 2 divides Lm   if and only if 3 divides m
(7) 3 divides Lm   if and only if m ≡ 2 (mod 4)
(8)F-n = (-1)n-1 Fn
(9)L-n = (-1)n-1 Ln
(10)Lk ≡ 3 (mod 4)   if 2 divides k and 3 does not divide k
(11)Lm+2k ≡ - Lm (mod Lk)
(12)Fm+2k ≡ - Fm (mod Lk)
(13)Lm+12 ≡ Lm (mod 8)

THE MAIN THEOREMS

Theorem 1. If Ln = x2, then n = 1 or 3.

Proof. If n is even, (3) gives

Ln = y2 ± 2 ≠ x2.
If n ≡ 1 (mod 4), then L1 = 1, whereas if n ≠ 1 we can write n = 1 + 2·3r·k where k has the required properties, and then obtain by (11)
Ln ≡ - L1 = -1 (mod Lk)
and so Ln ≠ x2 since -1 is a non-residue of Lk by (10). Finally, if n ≡ 3 (mod 4) then n = 3 gives L3 = 22, whereas if n ≠ 3, we write as before n = 3 + 2·3r·k and obtain
Ln ≡ - L3 = -4 (mod Lk)
and again Ln ≠ x2.

This concludes the proof of Theorem 1.

Theorem 2. If Ln = 2 x2, then n = 0 or ±6.

Proof. If n is odd and Ln is even, then by (6) n ≡ ±3 (mod 12) and so, using (13) and (9),

Ln ≡ 4 (mod 8)
and so Ln ≠ 2 x2.

Secondly, if n ≡ 0 (mod 4), then n = 0 gives L2 = 2, whereas if n ≠ 0, n = 2·3r·k and so
2 Ln ≡ -2 L0 = -4 (mod Lk)
whence
2 Ln ≠ y2,   i.e., Ln ≠ 2 x2

Thirdly, if n ≡ 6 (mod 8), then n = 6 gives L6 = 2·32, whereas if n ≠ 6, n = 6 + 2·3r·k where now 4 divides into k, and 3 does not divide into k and so
2 Ln ≡ -2 L6 = -36 (mod Lk)
and again, -36 is a non-residue of Lk using (7) and (10). Thus as before Ln ≠ 2 x2.

Finally, if n ≡ 2 (mod 8), then by (9) L-n = Ln where now -n ≡ 6 (mod 8) and so the only admissible value is -n = 6, i.e. n = -6.

This concludes the proof of Theorem 2.

Theorem 3. If Fn = x2, then n = 0, ±1, 2 or 12.

Proof. If n ≡ 1 (mod 4), then n = 1 gives F1 = 1, whereas if n ≠ 1, n = 1 + 2·3r·k and so

Fn ≡ - F1 = -1 (mod Lk)
whence Fn ≠ x2. If n ≡ 3 (mod 4), then by (8) F-n = Fn and -n ≡ 1 (mod 4) and as before we get only n = -1. If n is even, then by (1) Fn = F½n L½n and so, using (4) and (5), we obtain, if Fn = x2
either
3 divides n, F½n = 2 y2, L½n = 2 z2. By Theorem 2, the latter is possible only for ½n = 0, 6, or -6. The first two values also satisfy the former, while the last must be rejected because it does not.
or
3 does not divide n, F½n = y2, L½n = z2. By Theorem 1, the latter is possible only for ½n = 1 or 3, and again the second value must be rejected.

This concludes the proof of Theorem 3.

Theorem 4. If Fn = 2 x2, then n = 0, ±3, or 6.

Proof. If n ≡ 3 (mod 4), then n = 3 gives F3 = 2, whereas if n ≠ 3, n = 3 + 2·3r·k and so

2 Fn ≡ -2 F3 = -4 (mod Lk)
and so Fn ≠ 2 x2. If n ≡ 1 (mod 4), then as before F-n = Fn and we get only n = -3. If n is even, then since Fn = F½n L½n we must have if F½n = 2 x2
either
F½n = y2, L½n = 2 z2; then by Theorems 2 and 3 we see that the only value which satisfies both of these is ½n = 0
or
F½n = 2 y2, L½n = z2; then by Theorem 1, the second of these is satisfied only for ½n = 1 or 3. But the former of these does not satisfy the first equation.

This concludes the proof of the theorem.

REFERENCES

  1. M. Wunderlich, On the non-existence of Fibonacci Squares, Maths. of Computation, 17 (1963), p. 455.

  2. J. H. E. Cohn, On Square Fibonacci Numbers, Proc. Lond. Maths. Soc., 39 (1964) to appear.

  3. G. H. Hardy and E. M. Wright, Introduction to Theory of Numbers, 3rd. Edition, O.U.P. 1954, p. 148 et seq.

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EDITORIAL NOTE

Brother U. Alfred cheerfully acknowledges the priority of the essential method, used in "Lucas Squares" in the last issue of the Fibonacci Quarterly Journal, rest solely with J. H. E. Cohn. This was written at the request of the Editor and the unintentional omission of due credit rest solely with the Editor.