What are possible remainders when dividing with 9 squares of natural numbers?

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- Apr 26th 2006, 05:01 PMOReillyPossible remainders
What are possible remainders when dividing with 9 squares of natural numbers?

- Apr 26th 2006, 07:15 PMThePerfectHackerQuote:

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 - Apr 27th 2006, 02:27 AMOReilly
I must say I didn't think of that!

This problem has destroyed me! Whole day I couldn't solve it. Problems with divisibility are giving me most hard times and usually I can't solve them!