Order of a+1 When a has order 4 Modulo p

If  
\[2 \le a \lt p\]
  has order 3 modulo  
\[p\]
  then  
\[a^3 \equiv 1 \; (mod \; p) \rightarrow a^3 -1 \equiv 0 \; (mod \; p)\]
.  
\[a^3-1\]
  factorises as  
\[a^3-1=(a-1)(a^2+a+1)\]
  so  
\[a^3 -1 =(a-1)(a^2+a+1) \equiv 0 \; (mod \; p)\]
.  
\[gcd(a-1,p)=1\]
  so we can cancel the factor  
\[a-1\]
  and get  
\[(a^2+a+1) \equiv 0 \; (mod \; p)\]
.
Then  
\[-a^2 \equiv a+1 \; (mod \; p)\]
. We have to raise the left hand side of this congruence to the power of 6 to obtain 1 modulo  
\[p\]
, so  
\[a+1\]
  has order 6 modulo  
\[p\]
.