Question 110
Considering $\pmod{4}$ or otherwise, show that if $a, b, c, d$ are integers such that
$$
(3a + 5b)(7b + 11c)(13c + 17d)(19d + 23a) = 2001^{2001},
$$
then $a$ must be even.
$$
(3a + 5b)(7b + 11c)(13c + 17d)(19d + 23a) = 2001^{2001},
$$
then $a$ must be even.