Prove $A_q(3, 2) = q^2$ for all $q\ge 2$
I've tried using the sphere packing bound to find a max M for a (3, M, 2) code (with hamming distance 1), and found $M< {q^3\over 4q-3} < q^2$. I then tried the sphere packing bound again for some (3+1, M, 2+1) code and found $M \le {q^4 \over 4q - 3} > q^2$. My professor hasn't shown us any other method of calculating maximum M so I would greatly appreciate some guidance (Not necessarily the answer)
Edit: Solution
Obtaining an upper bound for $A_q(3, 2)$: After @Jyrki Lahtonen told me about the Singleton bound, I found here that for some q-ary (3, M, 2)-code you can obtain a maximum bound for M using the inequality $$M \le q^{n-d+1}$$ In this case, we see that $M \le q^{3-2+1} = q^2$ so $A_q(3, 2) \le q^2$
Obtaining a q-ary $(3, q^2, 2)$ code: I started by declaring the $(2, q^2, 1)$ code $C = F^2$ and adding a checksum digit to each codeword of C (since the distance of C is 1). The result is a new code $C_1=(3,q^2, 2)$. By finding $C_1$, I've proven $A_q(3, 2) \ge q^2$. Thus $A_q(3, 2) = q^2$.