Can't provide a proof, but using the computer for a and b,
each up to 1000, I obtained these four solutions -
EDIT: Sorry, first said a million, but it was only a thousand.
(a, b, k) =
(13, 5, 231) .......... k = 3 * 77
(34, 13, 1425) .......... k = 3 * 475
(233, 89, 62907) .......... k = 3 * 20969
(610, 233, 428217) .......... k = 3 * 142739
Still looking into it!!!
EDIT 1:
I seem to have detected a pattern in Fibonacci numbers -
check out the values of a and b.
Must look into that more closely, but for now, I have these
equations for a, b and k, each dependent on 'n' only,
for n = 1, 2, 3, ...
a = 89n^2 - 246n + 170
b = 34x^2 - 94n + 65
k = 233n^2 - 644n + 442
If everything is going smoothly there, it appears
there may be an infinite number of solutions.
EDIT 2:
Oh! Drat!, we ran out of Fibonacci numbers at n = 5.
But I think my equations are at fault, because the
pattern of Fibonacci numbers does seem to hold.
For a = 4181 and b = 1597, k = 20043711.
For a = 10946 and b = 4181, k = 137328513.
I'll have another go.
EDIT 3 :
So if the Fibonacci numbers are :
F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ..., then it
appears that there may be two patterns for a and b :
First pattern :
a = F(7) and b = F(5)
a = F(13) and b = F(11)
a = F(19) and b = F(17)
etc.
In general, a = F(6n + 1) and b = F(6n - 1).
Second pattern :
a = F(9) and b = F(7)
a = F(15) and b = F(13)
a = F(21) and b = F(19)
etc.
In general, a = F(6n + 3) and b = F(6n + 1).
EDIT 4:
Perhaps someone adept at Fibonacci identities
can verify those. It seems like a big undertaking.
But with : a³b + ab³ + a³ + ab² - a² - b² + (1 - k)ab + a - 1 = 0,
which can be transformed to :
k = a² + b² + b + 1 + [(a -1)(a² + 1) - b²] / (ab),
it's only necessary to verify that the fractional term is integral.
From there, two formulae for k should follow.
I therefore predict an infinite number of solutions.
EDIT 5:
Following is a list of Fibonacci numbers with the
pattern for each 'a' and 'b' and a list of 'k' values
I've confirmed, up to near the limit of my computer.
F(0) = 0
F(1) = 1
F(2) = 1
F(3) = 2
F(4) = 3
F(5) = 5 ---> b1
F(6) = 8
F(7) = 13 --> a1 and b2
F(8) = 21
F(9) = 34 --> a2
F(10) = 55
F(11) = 89 --> b3
F(12) = 144
F(13) = 233 --> a3 and b4
F(14) = 377
F(15) = 610 --> a4
F(16) = 987
F(17) = 1597 --> b5
F(18) = 2584
F(19) = 4181 --> a5 and b6
F(20) = 6765
F(21) = 10946 --> a6
F(22) = 17711
F(23) = 28657 --> b7
F(24) = 46368
F(25) = 75025 --> a7 and b8
F(26) = 121393
F(27) = 196418 --> a8
F(28) = 317811
F(29) = 514229 --> b9
F(30) = 832040
F(31) = 1346269 --> a9 and b10
F(32) = 2178309
F(33) = 3524578 --> a10
k1 = 231
k2 = 1425
k3 = 62907
k4 = 428217
k5 = 20043711
k6 = 137328513
k7 = 6450199347
k8 = 44209370601
k9 = 2076875723607
k10 = 14235100872177
EDIT 6:
Yippee! I got my proof that the fractional term above is
an integer for both patterns, in one fell swoop, with many
thanks to steiner1745. Reference :
http://au.answers.yahoo.com/question/index;_ylt=AgHUBADH6ewbSyUnRgfou.Lg5gt.;_ylv=3?qid=20080620235511AAueyUo As for nice, simple formulae for 'k', in terms of Phi, I tried.
For example, for the first pattern, I was successful in getting
a very long and tedious formula, that is not pretty enough to
display. A difficult simplification when you see that :
a = [Ø^(12n+2) + 1] / [Ø^(6n+1)√5]
and
b = [Ø^(12n-2) + 1] / [Ø^(6n-1)√5]
If you feel like doing a good job with that, it's all yours.