BMO 1 1998 question 2

Question

Let $a_1=19$, $a_2=98$. For $n\ge1$, define $a_{n+2}$ to be the remainder of $a_n + a_{n+1}$ when it is divided by $100$. What is the remainder when $a_1^2+a_2^2+\cdots+a_{1996}^2$ is divided by 8? I got (working mod 8) that the terms of the sequence squared are congruent to ${1,4,1,1,0,1}$ in this specific order, I also proved that $a_{n+1}^2$ is congruent to $(a_n + a_{n-1})^2$ from this I proved that if two consecutive terms of the sequence squared are congruent to 1 and 4 respectively then the the 3rd term squared is congruent to 1 and the fourth term squared is congruent to 1 from this I have reached the conclusion that the fifth term squared term is congruent to 0 or 4 and I think to complete my proof I need to show that $a_{6n}$ is congruent to $a_{6n+1}$, how do I do this? This is from BMO 1 1998 question 2.