MHB How can we show that one of these congruences hold?

  • Thread starter Thread starter mathmari
  • Start date Start date
AI Thread Summary
The discussion focuses on proving that for all integers squared, the results modulo 8 yield specific congruences: 0, 1, or 4. The set M is defined as the squares of integers, and examples are provided to illustrate the congruences. Participants analyze the periodicity of the remainders when squaring integers, noting a repeating pattern every four integers. The proof is structured by considering even and odd integers separately, demonstrating that the squares of both types yield the required congruences. The conversation emphasizes the mathematical reasoning behind these findings.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

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)
 
Mathematics news on Phys.org
mathmari said:
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)
 
Klaas van Aarsen said:
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 $0,1,4,9=1$.
 
Last edited by a moderator:
mathmari said:
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.

mathmari said:
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)
 
mathmari said:
Hey! :o

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.
 
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 n^2= 4m^2.
m is either even or odd
1a)Suppose m is even. Then m= 2k and n^2= 4(4k^2)= 8(2k^2) and n^2 is congruent to 0 mod 8.
1b)Suppose m is odd. Then m= 2k+ 1 and n^2= 4(2k+1)^2= 4(4k^2+ 4k+ 1)= 8(2k^2+ 2k)+ 4 and n^2 is congruent to 4 mod 8.

2) Suppose n is odd. Then n= 2m+ 1 for some integer m so n^2= 4m^2+ 4m+ 1= 4(m^2+ m)+ 1.
2a) Suppose m is even. Then m= 2k so n^2= 4(4k^2+ 2k)+ 1= 8(2k^2+ k)+ 1 and n^2 is congruent to 1 mod 8.
2b) Suppose m is odd. Then m= 2k+ 1 so n^2= 4(4k^2+ 4k+ 1+2k+ 1)+ 1= 4(4k^2+ 6k+ 2)+ 1= 8(2k^2+ 3k+ 1)+ 1 and n^2 is congruent to 1 mod 8.
 
Last edited by a moderator:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top