Fibonacci proof by Strong Induction Prove by strong induction that for a ∈ A we have 
$F_a + 2F_{a+1} = F_{a+4} − F_{a+2}.$
$F_a$ is the $a$'th element in the Fibonacci sequence

Do you consider the sequence starting at 0 or 1? I will assume 1.
If that is the case, $F_{a+1} = F_a + F_{a-1}) $ for all integers where $a \geq 3$. 
The original equation states $F_{a+1} = (F_a) + F_{a-1} $. 
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$F_{a+1} = (F_a) + F_{a-1} $
$-(F_a) = -F_{a+1}+ F_{a-1} $
$F_a = F_{a+1}- F_{a-1}$. This equation is important.
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$F_{a+3} = F_{a+4} - F_{a+2}$   
after subtracting and dividing by -1 we have
$F_{a+4} = F_{a+3} + F_{a+2}$. This equation is important too.
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By shifting we have $F_{a+3} = F_{a+2} + F_{a+1}$ and $F_{a+2} = F_{a+1} + F_{a}$. These formulas will be used to "reduce the power," in a sense.
$F_{a+4} - F_{a+2} = F_{a+2} + F_{a+1} + F_{a+2} - F_{a+2}$   
$F_{a+4} - F_{a+2} = F_{a+2} + F_{a+1}$  
By using the substitution $F_{a+2} = F_{a+1} + F_{a}$ we have $F_{a+4} - F_{a+2} = (F_{a} + F_{a+1}) + F_{a+1}$  
Therefore $F_{a+4} - F_{a+2} = F_{a} + 2F_{a+1}$  
