<s>[INST] The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square. 
The 4 digit  base $6$ number $abcd$ with $a>0$ and $d$ odd  is a perfect square. List all possible values of $c$. (The letters are the digits of the base 6 number.)

I've rewritten $abcd_6$ into $216a + 36b + 6c + d$, and I know that $d$ can be $1,3,5$, but now I'm stuck. I guess I could start plugging in some numbers since $a,b,c,d \le 5$, but I feel like there would be a more efficient way. 
 [/INST] Work modulo $36$, test only the odd numbers $15$ through $35$:
$^\text{(we may omit $1$ through $13$, since these would not satisfy $a>0$)}$
$$\,$$
$$(15)^2 \equiv 9 \pmod{36} $$
$$(17)^2 \equiv 1 \pmod{36} $$
$$(19)^2 \equiv 1 \pmod{36}$$
$$(21)^2 \equiv 9 \pmod{36} $$
$$(23)^2 \equiv 25 \pmod{36}$$
$$(25)^2 \equiv 13 \pmod{36} $$
$$(27)^2 \equiv 9 \pmod{36} $$
$$(29)^2 \equiv 13 \pmod{36} $$
$$(31)^2 \equiv 25 \pmod{36}$$
$$(33)^2 \equiv 9 \pmod{36} $$
$$(35)^2 \equiv 1 \pmod{36} $$
$$\,$$
The possible remainders for odd perfect squares, modulo $36$, are $1$, $9$, $13$, and $25$. 

In base $6$, these are equivalent to the last two digits being $$01_6,\,13_6,\,21_6,\,41_6$$
In particular, $$\boxed{\,c \, \in \, \left\{0,1,2,4\right\}\,\,}$$
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