, 2015

Taking mod 4, we can find that
$$(a_n, a_{n+1}) = (0, 3) \pmod 4$$
$$(a_n, a_{n+1}) = (1, 0) \pmod 4$$
$$(a_n, a_{n+1}) = (2, 3) \pmod 4$$
$$(a_n, a_{n+1}) = (3, 2) \pmod 4.$$
Note that a perfect square modulo 4 can only be 0 or 1. If $a_n = 0 \pmod 4$, then for all $m > n$, we must have $a_m = 2 \pmod 4$ or $a_m = 3 \pmod 4$, so subsequent terms cannot be perfect squares. And if $a_n = 1 \pmod 4$, then $a_{n+1} = 0 \pmod 4$, and the previous discussion still applies. From this we know that at most two terms $(a_n, a_{n+1})$ can be perfect squares.
Now suppose $a_n = p^2, a_{n+1} = q^2$. Substituting into the original equation, we get $p^6 + 103 = q^2$, that is, $103 = (q + p^3)(q - p^3)$. Since 103 is prime, we must have $q + p^3 = 103$ and $q - p^3 = 1$, that is, $p^3 = 51$, a contradiction! Therefore, at most one term can be a perfect square.