When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.
And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.
Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.
$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256
$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256
Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)
$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, tn = a + (n - 1)d