Converted to HTML 4.0 by Christopher Carl Heckman
I have taken some liberties with the paper, for the sake of universality of browsers. For instance, statements like "4 ¦ k" have been written out as "4 divides into k".
This paper originally appeared as:
J. H. E. Cohn, "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 1964, pp. 109-113.
y^{2} = D x^{4} + 1to be completely solved for certain values of D. Of course the Fibonacci case is simply D = 5.
(1) | 2 F_{m+n} | = F_{m} L_{n} + F_{n} L_{m} |
(2) | 2 L_{m+n} | = 5 F_{m} F_{n} + L_{m} L_{n} |
(3) | L_{2m} | = L_{m}^{2} + (-1)^{m-1} 2 |
(4) | (F_{3m}, L_{3m}) | = 2 |
(5) | (F_{n}, L_{n}) | = 1 if 3 does not divide n |
(6) | 2 divides L_{m} if and only if 3 divides m | |
(7) | 3 divides L_{m} if and only if m ≡ 2 (mod 4) | |
(8) | F_{-n} | = (-1)^{n-1} F_{n} |
(9) | L_{-n} | = (-1)^{n-1} L_{n} |
(10) | L_{k} | ≡ 3 (mod 4) if 2 divides k and 3 does not divide k |
(11) | L_{m+2k} | ≡ - L_{m} (mod L_{k}) |
(12) | F_{m+2k} | ≡ - F_{m} (mod L_{k}) |
(13) | L_{m+12} | ≡ L_{m} (mod 8) |
Proof. If n is even, (3) gives
L_{n} = y^{2} ± 2 ≠ x^{2}.If n ≡ 1 (mod 4), then L_{1} = 1, whereas if n ≠ 1 we can write n = 1 + 2·3^{r}·k where k has the required properties, and then obtain by (11)
L_{n} ≡ - L_{1} = -1 (mod L_{k})and so L_{n} ≠ x^{2} since -1 is a non-residue of L_{k} by (10). Finally, if n ≡ 3 (mod 4) then n = 3 gives L_{3} = 2^{2}, whereas if n ≠ 3, we write as before n = 3 + 2·3^{r}·k and obtain
L_{n} ≡ - L_{3} = -4 (mod L_{k})and again L_{n} ≠ x^{2}.
This concludes the proof of Theorem 1.
Theorem 2. If L_{n} = 2 x^{2}, then n = 0 or ±6.
Proof. If n is odd and L_{n} is even, then by (6) n ≡ ±3 (mod 12) and so, using (13) and (9),
L_{n} ≡ 4 (mod 8)and so L_{n} ≠ 2 x^{2}.
2 L_{n} ≡ -2 L_{0} = -4 (mod L_{k})whence
2 L_{n} ≠ y^{2}, i.e., L_{n} ≠ 2 x^{2}
2 L_{n} ≡ -2 L_{6} = -36 (mod L_{k})and again, -36 is a non-residue of L_{k} using (7) and (10). Thus as before L_{n} ≠ 2 x^{2}.
This concludes the proof of Theorem 2.
Theorem 3. If F_{n} = x^{2}, then n = 0, ±1, 2 or 12.
Proof. If n ≡ 1 (mod 4), then n = 1 gives F_{1} = 1, whereas if n ≠ 1, n = 1 + 2·3^{r}·k and so
F_{n} ≡ - F_{1} = -1 (mod L_{k})whence F_{n} ≠ x^{2}. If n ≡ 3 (mod 4), then by (8) F_{-n} = F_{n} and -n ≡ 1 (mod 4) and as before we get only n = -1. If n is even, then by (1) F_{n} = F_{½n} L_{½n} and so, using (4) and (5), we obtain, if F_{n} = x^{2}
3 divides n, F_{½n} = 2 y^{2}, L_{½n} = 2 z^{2}. 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, F_{½n} = y^{2}, L_{½n} = z^{2}. 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 F_{n} = 2 x^{2}, then n = 0, ±3, or 6.
Proof. If n ≡ 3 (mod 4), then n = 3 gives F_{3} = 2, whereas if n ≠ 3, n = 3 + 2·3^{r}·k and so
2 F_{n} ≡ -2 F_{3} = -4 (mod L_{k})and so F_{n} ≠ 2 x^{2}. If n ≡ 1 (mod 4), then as before F_{-n} = F_{n} and we get only n = -3. If n is even, then since F_{n} = F_{½n} L_{½n} we must have if F_{½n} = 2 x^{2}
F_{½n} = y^{2}, L_{½n} = 2 z^{2}; then by Theorems 2 and 3 we see that the only value which satisfies both of these is ½n = 0or
F_{½n} = 2 y^{2}, L_{½n} = z^{2}; 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.