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- #1

- Apr 14, 2013

- 3,958

We define the set $M:=\{x^2\mid x\in\mathbb{Z}\}$.

- Give 8 different elements of M.
- Give for each element of question 1, m, two numbers $q\in \mathbb{Z}, r\in \{0,1,\ldots , 7\}$ such that $m=8q+r$.
- Show that for all $x\in M$ one of the following holds:

$x\overset{(8)}{\equiv}0$, $x\overset{(8)}{\equiv}1$, $x\overset{(8)}{\equiv}4$

I have done the following:

- We have $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64$.
- $1=8\cdot 0+1, 4=8\cdot 0+4, 9=8\cdot 1+1, 16=8\cdot 2+0, 25=8\cdot 3+1, 36=8\cdot 4+4, 49=8\cdot 6+1, 64=8\cdot 8+0$
- How can we show that?