Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28 This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents. 
Prove that if an integer $a$ is both a square and a cube then $a  \equiv 0,1,9, \textrm{ or } 28  (\textrm{ mod}\ 36)$ 
An outline of the proof I have is
Any such integer $a$ has $a = x^2$ and $a = y^3$ for some integers $x,y$
Then by the Division Algorithm, $x = 36s + b$ for some integers $s,b$ with $0 \le b \lt 36$ and $y = 36t + c$ for some integers $t,c$ with $0 \le c \lt 36$
Using binomial theorem, it is easy to show that $x^2 \equiv b^2$ and $y^3 \equiv c^3$
Then $a \equiv b^2$ and $a \equiv c^3$
By computer computation (simple script), the intersection of the possible residuals for any value of $b$ and $c$ in the specified interval is 0,1,9,28 
These residuals are possible but not actual without inspection which shows $0^2 = 0^3 \equiv 0$ , $1^2 = 1^3 \equiv 1$ , $27^2 = 9^3 \equiv 9$, and $8^2 = 4^3 \equiv 28$ $\Box$
There is surely a more elegant method, can anyone hint me in the right direction.

What you did is correct, but yes, a lot of the work (especially the computer check) could have been avoided.
Firstly, if $a$ is both a square and a cube, then it is a sixth power.
This is because, for any prime $p$, $p$ divides $a$ an even number of times (since it is a square), and a multiple of 3 number of times (since it is a cube), so $p$ divides $a$ a multiple of 6 number of times altogether, and since this is true for any prime $p$, $a$ is a perfect sixth power.
So write $a = z^6$.
Next, rather than working mod $36$, it will be nice to work mod $9$ and mod $4$ instead; this is equivalent by the chinese remainder theorem.
So:


*

*Modulo $9$, $z^6 \equiv 0 \text{ or } 1$. You can see this just by checking every integer or by applying the fact that $\varphi(9) = 6$.

*Modulo $4$, $z^6 \equiv 0 \text{ or } 1$. This is easy to see; $0^6 = 0$, $1^6 = 1$, $(-1)^6 = 1$, and $2^6 \equiv 0$.
So $a = z^6$ is equivalent to $0$ or $1$ mod $4$ and mod $9$.
By the chinese remainder theorem, this gives four possibilities:


*

*$a \equiv 0 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 0 \pmod{36}$

*$a \equiv 0 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 28 \pmod{36}$

*$a \equiv 1 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 9 \pmod{36}$

*$a \equiv 1 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 1 \pmod{36}$.
