proof that $\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer I would appreciate if somebody could help me with the following problem: 
Q: How to proof? 
If $\{a_n\}$ satisfy $a_{1}=a$, $a_2=b$, $a_{n+2}=a_{n+1}+a_{n}$($a,b$: positive integers)
then proof that 
$\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer
I try  start by mathematical induction but....,
find $f(n)=\frac{a_{4n}-a_2}{a_{2n+1}}$ $f(1)=1$, $f(2)=4$, $f(3)=11$, $f(4)=29$,...

$R=\frac{a_{4n}-a_2}{a_{2n+1}}$
$a_1=a$
$a_2=b$
$a_3=a+b$
$a_4=a+2b$
$a_5=2a+3b$
$a_6=3a+5b$
$a_7=5a+8b$
$a_8=8a+13b$
For n=2 we have:
$\frac{8a+13b-b}{2a+3b}=4$
By experimental induction if R is true for n=2 it must also be true for n+1=3, we check this:
$\frac{a_{12}-b}{a_7}=\frac{55a+89b-b}{5a+8b}=11$ 
This true for any value of a and b,for example $a_{12}=65$ for a=0, b=1 and n=3 :
$\frac{65-1}{8}=8$
