What are possible remainders when dividing with 9 squares of natural numbers?
You can attack the problem with Division Algorithm.Originally Posted by OReilly
Meaning any number must be exactly in one of these forms,
$\displaystyle 9k$
$\displaystyle 9k+1$
$\displaystyle 9k+2$
.........................
$\displaystyle 9k+8$
When you square then, you get, the only possibilities,
$\displaystyle 9k$
$\displaystyle 9k+1$
$\displaystyle 9k+4$
$\displaystyle 9k+7$
Thus, remainders are 0,1,4,7