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Number Theory How can we show that one of these congruences hold?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,958
Hey!! :eek:

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

  1. Give 8 different elements of M.
  2. Give for each element of question 1, m, two numbers $q\in \mathbb{Z}, r\in \{0,1,\ldots , 7\}$ such that $m=8q+r$.
  3. Show that for all $x\in M$ one of the following holds:
    $x\overset{(8)}{\equiv}0$, $x\overset{(8)}{\equiv}1$, $x\overset{(8)}{\equiv}4$


I have done the following:
  1. We have $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64$.
  2. $1=8\cdot 0+1, 4=8\cdot 0+4, 9=8\cdot 1+1, 16=8\cdot 2+0, 25=8\cdot 3+1, 36=8\cdot 4+4, 49=8\cdot 6+1, 64=8\cdot 8+0$
  3. How can we show that? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,593
Leiden
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? (Thinking)

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 $y^2\pmod 8$? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,958
Suppose we write $y=4k+r$ for some integer $k$ and $r\in\{0,1,2,3\}$.
What will be the possibilities for $y^2\pmod 8$? (Wondering)
Why do we take the y in this form? Because it is periodic of period 4? (Wondering)

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 $0,1,4,9=1$.
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,593
Leiden
Why do we take the y in this form? Because it is periodic of period 4?
Yep. (Nod)
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.

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? (Wondering)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Hey!! :eek:

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

  1. Give 8 different elements of M.
  2. Give for each element of question 1, m, two numbers $q\in \mathbb{Z}, r\in \{0,1,\ldots , 7\}$ such that $m=8q+r$.
  3. Show that for all $x\in M$ one of the following holds:
    $x\overset{(8)}{\equiv}0$, $x\overset{(8)}{\equiv}1$, $x\overset{(8)}{\equiv}4$


I have done the following:
  1. We have $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64$.
  2. $1=8\cdot 0+1, 4=8\cdot 0+4, 9=8\cdot 1+1, 16=8\cdot 2+0, 25=8\cdot 3+1, 36=8\cdot 4+4, 49=8\cdot 6+1, 64=8\cdot 8+0$
  3. How can we show that? (Wondering)
Look at your answers to (2)! Each has 8 times some number plus 1, 4, 1, 0, 1, 4, 1, 0.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Tedious- we need to look at four "cases".

Every number in Z is either even or odd.
1) Suppose n is even. Then n= 2m for some integer m so [tex]n^2= 4m^2[/tex].
m is either even or odd
1a)Suppose m is even. Then m= 2k and [tex]n^2= 4(4k^2)= 8(2k^2)[/tex] and [tex]n^2[/tex] is congruent to 0 mod 8.
1b)Suppose m is odd. Then m= 2k+ 1 and [tex]n^2= 4(2k+1)^2= 4(4k^2+ 4k+ 1)= 8(2k^2+ 2k)+ 4[/tex] and [tex]n^2[/tex] is congruent to 4 mod 8.

2) Suppose n is odd. Then n= 2m+ 1 for some integer m so [tex]n^2= 4m^2+ 4m+ 1= 4(m^2+ m)+ 1[/tex].
2a) Suppose m is even. Then m= 2k so [tex]n^2= 4(4k^2+ 2k)+ 1= 8(2k^2+ k)+ 1[/tex] and [tex]n^2[/tex] is congruent to 1 mod 8.
2b) Suppose m is odd. Then m= 2k+ 1 so [tex]n^2= 4(4k^2+ 4k+ 1+2k+ 1)+ 1= 4(4k^2+ 6k+ 2)+ 1= 8(2k^2+ 3k+ 1)+ 1[/tex] and [tex]n^2[/tex] is congruent to 1 mod 8.
 
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