## Fibonacci num3ers : a surprising occurrence

To find positive integers   $(a, \; b, \; c)$   such that

$a \, b \; + \; a \; + \; b$
$b \, c \; + \; b + \; c$
$c \, a \; + \; c + \; a$
$a \, b \; + \; c$
$a \, c \; + \; b$

are all squares

that is,

$a \, b \; + \; a \; + \; b \; = \; p^2$
$b \, c \; + \; b + \; c \; = \; q^2$
$c \, a \; + \; c + \; a \; = \; r^2$
$a \, b \; + \; c \; = \; s^2$
$a \, c \; + \; b \; = \; t^2$

Note that

$1 \; + \; p^2 \; = \; (a + 1) \,(b + 1)$
$1 \; + \; q^2 \; = \; (b + 1) \,(c + 1)$
$1 \; + \; r^2 \; = \; (a + 1) \,(c + 1)$

It happens that the first few Fibonacci numbers can be used to solve this system of equations.

Fibonacci numbers:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, …

$F_2 = 1$,     $F_3 = 2$,     $F_4 = 3$

$1^2 \; + \; 1 \; = \; (a + 1) \,(b + 1)$
a = 0,   b = 1             a = 1,   b = 0

$2^2 \; + \; 1 \; = \; (a + 1) \,(c + 1)$
a = 0,   c = 4             a = 4,   c = 0

$3^2 \; + \; 1 \; = \; (b + 1) \,(c + 1)$
b = 0,   c = 9             b = 9,   c = 0
b = 1,   c = 4             b = 4,   c = 1

So the solution is :    a = 0,    b = 1,    c = 4

$a \, b \; + \; a \; + \; b \; = \; (0\times 1) \; + \; 0 + \; 1 \; = 1 \; = \; 1^2$
$b \, c \; + \; b \; + \; c \; = \; (1\times 4) \; + \; 1 + \; 4 \; = 9 \; = \; 3^2$
$c \, a \; + \; c \; + \; a \; = \; (4\times 0) \; + \; 4 + \; 0 \; = 4 \; = \; 2^2$
$a \, b \; + \; c \; = \; (0\times 1) \; + \; 4 \; = \; 2^2$
$a \, c \; + \; b \; = \; (0\times 4) \; + \; 1 \; = \; 1^2$

$F_4 = 3$     $F_5 = 5$     $F_6 = 8$

$3^2 \; + \; 1 \; = \; (a + 1) \,(b + 1)$
a = 0,   b = 9,     a = 9,   b = 0
a = 1,   b = 4,     a = 4,   b = 1

$5^2 \; + \; 1 \; = \; (a + 1) \,(c + 1)$
a = 0,   c = 25,     a = 25,   c = 0
a = 1,   c = 12,     a = 12,   c = 1

$8^2 \; + \; 1 \; = \; (b + 1) \,(c + 1)$
b = 0,   c = 64,     b = 64,   c = 0
b = 4,   c = 12,     b = 12,   c = 4

So the solution is :    a = 1,    b = 4,    c = 12

$a \, b \; + \; a \; + \; b \; = \; (1\times 4) \; + \; 1 \; + \; 4 \; = \; 9 \; = \; 3^2$
$b \, c \; + \; b \; + \; c \; = \; (4\times 12) \; + 4 \; + \; 12 \; = \; 64 \; = \; 8^2$
$c \, a \; + \; c \; + \; a \; = \; (12\times 1) \; + \; 12 \; + \; 1 \; = \; 25 = \; 5^2$
$a \, b \; + \; c \; = \; (1\times 4) \; + \; 12 \; = \; 16 \; = \; 4^2$
$a \, c \; + \; b \; = \; (1\times 12) \; + \; 4 \; = \; 16 \; = \; 4^2$

$F_6 = 8$     $F_7 = 13$     $F_8 = 21$

$8^2 \; + \; 1 \; = \; (a + 1) \,(b + 1)$
a = 0,   b = 64,     a = 64,   b = 0
a = 4,   b = 12,     a = 12,   b = 4

$13^2 \; + \; 1 \; = \; (a + 1) \,(c + 1)$
a = 0,   c = 169,     a = 169,   c = 0
a = 1,   c = 84,     a = 84,   c = 1
a = 4,   c = 33,     a = 33,   c = 4
a = 9,   c = 16,     a = 16,   c = 9

$21^2 \; + \; 1 \; = \; (b + 1) \,(c + 1)$
b = 0,   c = 441,     b = 441,   c = 0
b = 1,   c = 220,     b = 220,   c = 1
b = 12,   c = 33,     b = 33,   c = 12
b = 16,   c = 25,     b = 25,   c = 16

So,     a = 4,    b = 12,    c = 33

$a \, b \; + \; a \; + \; b \; = \; (4\times 12) \; + \; 4 \; + \; 12 \; = \; 64 \; = \; 8^2$
$b \, c \; + \; b + c \; = \; (12\times 33) \; + \; 12 \; + \; 33 \; = \; 441 \; = \; 21^2$
$c \, a \; + \; c + a \; = \; (33\times 4) \; + \; 33 \; + \; 4 \; = \; 169 \; = \; 13^2$
$a \, b \; + \; c \; = \; (4\times 12) \; + \; 33 \; = \; 81 \; = \; 9^2$
$a \, c \; + \; b \; = \; (4\times 33) \; + \; 12 \; = \; 144 \; = \; 12^2$

$F_8 = 21$     $F_9 = 34$     $F_{10} = 55$

$21^2 \; + \; 1 \; = \; (a + 1)(b + 1)$
a = 0,   b = 441,     a = 441,   b = 0
a = 1,   b = 220,     a = 220,   b = 1
a = 12,   b = 33,     a = 33,   b = 12
a = 16,   b = 25,     a = 25,   b = 16

$34^2 \; + \; 1 \; = \; (a + 1)(c + 1)$
a = 0,   c = 1156,     a = 1156,   c = 0
a = 12,   c = 88,     a = 88,   c = 12

$55^2 \; + \; 1 \; = \; (b + 1)(c + 1)$
b = 0,   c = 3025,     b = 3025,   c = 0
b = 1,   c = 1512,     b = 1512,   c = 1
b = 16,   c = 177,     b = 177,   c = 16
b = 33,   c = 88,     b = 88,   c = 33

So,     a = 12,    b = 33,    c = 88

$a \, b \; + \; a \; + \; b \; = \; (12\times 33) \; + \; 12 \; + \; 33 \; = \; 441 \; = \; 21^2$
$b \, c \; + \; b \; + \; c \; = \; (33\times 88) \; + \; 33 \; + \; 88 \; = \; 3025 \; = \; 55^2$
$c \, a \; + \; c \; + \; a \; = \; (88\times 12) \; + \; 88 \; + \; 12 \; = \; 1156 \; = \; 34^2$
$a \, b \; + \; c \; = \; (12\times 33) \; + \; 88 \; = \; 484 \; = \; 22^2$
$a \, c \; + \; b \; = \; (12\times 88) \; + \; 33 \; = \; 1089 \; = \; 33^2$

Does there exist an infinity of Fibonacci triplets satisfying the given conditions?