<s>[INST] Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$ Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$
My Proof:
Ten cases exist, yielding the following equalities:
$$(1\mod{10})^2 = 1\mod{10}$$
$$(2\mod{10})^2 = 4\mod{10}$$
$$(3\mod{10})^2 = 9\mod{10}$$
$$(4\mod{10})^2 = 6\mod{10}$$
$$(5\mod{10})^2 = 5\mod{10}$$
$$(6\mod{10})^2 = 6\mod{10}$$
$$(7\mod{10})^2 = 9\mod{10}$$
$$(8\mod{10})^2 = 4\mod{10}$$
$$(9\mod{10})^2 = 1\mod{10}$$
$$(0\mod{10})^2 = 0\mod{10}$$
Since the proposition holds for all possible cases, the proposition holds.
Is this an acceptable proof for the proposition?
What is the simplest proof for this?
Note: This is not a homework question, just a question from a weekly tutorial sheet.
 [/INST] For a more "clever" solution (but yours looks perfect):
As $10=5\times 2$, let us consider 
$$
x^2 = 10a + 5b+ r
\\ 0\le 5b+r< 10
\\ 0\le r< 5
$$
then reducing modulo 5:
$$
r = x^2 \in \{ 0,1,4
\}\mod 5\\
5b+r \in \{ 0,1,4,0+5=5,1+5=6,4+5=9
\}\mod 5\\
$$
</s>