Solution with SuperJJ Consider this equation $\mod 5$. We see that for all $a$, we have $a^4 \equiv 1 \mod 5$, so $(a^4 - 1) \equiv 0 \mod 5$ so our expression is congruent to $0 \mod 5$ Now consider this equation $\mod 7$. If we want $(a^4 - 1) \equiv 0 \mod 7$, we must have $a^4 \equiv 1 \mod 7$ or $a \equiv 1 \mod 7$ and $a \equiv 6 \mod 7$. Looking at the $a^4+15a^2+1$ term, for that to be equal to $0 \mod 7$ we must have $a^4 + 15a^2 + 1$ equivalent to $2,3,4$ or $5 \mod 7$. Thus, our expression is congruent to $0 \mod 7$, so it is congruent to $0 \mod 35$ $\blacksquare$

Another Nice Solution: Factor the original expression into$$(a-1)(a+1)(a^2+1)((a^2+19)(a^2-4) + 77) = (a-1)(a+1)(a^2+1)((a^2+19)(a-2)(a+2)+77)$$From here, it's obvious that for any $a$ that is relatively prime with $35$, the expression is a multiple of 5. We now must check mod 7. The only possible integers $a$ must be in the form $7x+1, 7x+2, 7x+3, 7x+4, 7x+5,$ and/or $7x+6$. If we can prove that for each of these 6 cases, the expression becomes 0mod7, then we have finished the problem. Testing each case, we have: 1) $7x+1 \longrightarrow (a-1)$. We have an a-1 in our factored expression meaning if a is in the form 7x+1, it will become 7x and thus a multiple of 7. Adding that to 77 produces another multiple of 7 and so the expression is a multiple of 7. 2) $7x+2 \longrightarrow (a-2)$. If $a$ is in the form $7x+2$, plugging that in to our original expression, a 2 will be subtracted from it turning it into a multiple of 7 and just like 1), it will then become 0mod7. 3) $7x+3$. If we were to plug this into (a^2+19), we would square (7x+3) to obtain $49x^2 + 42x + 9$ and add 19 to obtain $49x^2 + 42x + 28$ which is clearly a multiple of 7. The other 3 cases are done in a similar fashion. Thus, we prove that it is a multiple of 7 for all possible cases of $a$ and a multiple of $5$ and we are done. $\blacksquare$ - Solution by SuperJJ's relative