
MHB Master
#1
February 6th, 2020,
01:58
Hey!!
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?

February 6th, 2020 01:58
# ADS
Circuit advertisement

MHB Seeker
#2
February 6th, 2020,
05:47
Originally Posted by
mathmari
We define the set $M:=\{x^2\mid x\in\mathbb{Z}\}$.
3. Show that for all $x\in M$ one of the following holds: $x\overset{(8)}{\equiv}0$, $x\overset{(8)}{\equiv}1$, $x\overset{(8)}{\equiv}4$
Hey mathmari!!
If $x\in M$, then there must be a $y\in\mathbb Z$ such that $x=y^2$ yes?
In question 2 we can see there seems to be a repeating pattern, don't we?
The remainers modulo 8 for consecutive values of $y$ are 1,4,1,0,1,4,1,0 after all.
That seem to be periodic with period 4, doesn't it?
Suppose we write $y=4k+r$ for some integer $k$ and $r\in\{0,1,2,3\}$.
What will be the possibilities for $y^2\pmod 8$?

MHB Master
#3
February 6th, 2020,
09:52
Thread Author
Originally Posted by
Klaas van Aarsen
Suppose we write $y=4k+r$ for some integer $k$ and $r\in\{0,1,2,3\}$.
What will be the possibilities for $y^2\pmod 8$?
Why do we take the y in this form? Because it is periodic of period 4?
The square is as follows:
$$y^2\mod 8\equiv 16k^2+8kr+r^2\mod 8\equiv r^2$$
The possible values of $r^2$ are $0,1,4,9=1$.
Last edited by mathmari; February 6th, 2020 at 12:57.

MHB Seeker
#4
February 6th, 2020,
14:06
Originally Posted by
mathmari
Why do we take the y in this form? Because it is periodic of period 4?
Yep.
It seems to be periodic with period 4, so we try to prove that it is indeed the case. And additionally what the possible remainders are.
Originally Posted by
mathmari
The square is as follows:
$$y^2\mod 8\equiv 16k^2+8kr+r^2\mod 8\equiv r^2$$
The possible values of $r^2$ are $0,1,4,9=1$.
So?

#5
February 7th, 2020,
08:11
Originally Posted by
mathmari
Hey!!
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?
Look at your answers to (2)! Each has 8 times some number plus 1, 4, 1, 0, 1, 4, 1, 0.