5. Given that among the four positive integers $a, b, c, d$, $a$ leaves a remainder of $1$ when divided by $9$, $b$ leaves a remainder of $3$ when divided by $9$, $c$ leaves a remainder of $5$ when divided by $9$, and $d$ leaves a remainder of $7$ when divided by $9$. Then the two numbers that cannot be perfect squares are ( ).
(A) $a, b$
(B) $b, c$
(C) $c, d$
(D) $d, a$
5.B.

Let the integer $x$ be divided by 9 with a remainder of $0, \pm 1, \pm 2, \pm 3, \pm 4$, then $x^{2}$ divided by 9 has a remainder of $0,1,4,7$. Therefore, positive integers that leave a remainder of $2,3,5,6,8$ when divided by 9 cannot be perfect squares, so $b$ and $c$ cannot be perfect squares.