Show that the last decimal digit of a perfect square cannot be 2,3,7, or 8.
Any advice? I don't know where to begin to show this...
Although you figured it out, just look to this proof (if yours is different of course) :
Here is a little table containing the last digit of any number , and the last digit of its square (thanks to Soroban for the array template) :
The last digits , , and are missing. Therefore, a perfect square cannot have , , or as a last digit.