## Fibonacci in arithmetic progression

Fibonacci numbers is defined by the recurrence relation:

$F_{n} \; = \; F_{n-1} \; + \; F_{n-2}$
$F_0 \; = \; 0$
$F_1 \; = \; 1$

then,   $F_{n+1} \; = \; F_{n} \; + \; F_{n-1}$

$F_{n} \; - \; F_{n-2} \; = \; F_{n-1}$
$F_{n+1} \; - \; F_{n} \; = \; F_{n-1}$

$F_{n} \; - \; F_{n-2} \; = \; F_{n+1} \; - \; F_{n} \; = \; F_{n-1}$

giving us all increasing arithmetic progressions formed of three terms of the Fibonacci sequence

For example,

$F_2 - F_0 \; = \; F_3 - F_2 \; = \; 1 - 0 \; = \; 2 - 1 \; = \; 1 \; = \; F_1 \; = \; F_2$

$F_3 - F_1 \; = \; F_4 - F_3 \; = \; 2 - 1 \; = \; 3 - 2 \; = \; 1 \; = \; F_1 \; = \; F_2$

$F_4 - F_2 \; = \; F_5 - F_4 \; = \; 3 - 1 \; = \; 5 - 3 \; = \; 2 \; = \; F_3$

$F_5 - F_3 \; = \; F_6 - F_5 \; = \; 5 - 2 \; = \; 8 - 5 \; = \; 3 \; = \; F_4$

$F_6 - F_4 \; = \; F_7 - F_6 \; = \; 8 - 3 \; = \; 13 - 8 \; = \; 5 \; = \; F_5$

$F_7 - F_5 \; = \; F_8 - F_7 \; = \; 13 - 5 \; = \; 21 - 13 \; = \; 8 \; = \; F_6$

$F_8 - F_6 \; = \; F_9 - F_8 \; = \; 21 - 8 \; = \; 34 - 21 \; = \; 13 \; = \; F_7$

$F_9 - F_7 \; = \; F_{10} - F_9 \; = \; 34 - 13 \; = \; 55 - 34 \; = \; 21 \; = \; F_8$

Prove that there are no increasing arithmetic progressions formed of four terms of this sequence.