How can we show that one of these congruences hold?

• MHB
• mathmari
In summary: Suppose n is even. Then n= 2m for some integer m so n^2= 4m^2.3a) Suppose m is odd. Then m= 2k so n^2= 4(4k^2+ 2k)+ 1= 8(2k^2+ k)+ 1 and n^2 is congruent to 1 mod 8.3b) Suppose m is even. 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
mathmari
Gold Member
MHB
Hey!

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)

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 $0,1,4,9=1$.

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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!

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.

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1. How do we determine if two shapes are congruent?

To show that two shapes are congruent, we can use several methods such as superposition, side-angle-side (SAS) congruence, angle-side-angle (ASA) congruence, and side-side-side (SSS) congruence. These methods involve comparing the corresponding sides and angles of the two shapes to see if they are equal in length and measure.

2. Can we prove congruence using only one pair of corresponding sides and angles?

No, to prove that two shapes are congruent, we need to have at least three pairs of corresponding sides and angles that are equal. This is because one pair of corresponding sides and angles is not enough to uniquely determine the shape and size of a figure.

3. How does the concept of congruence relate to similarity?

Congruence and similarity are both ways of comparing two shapes. However, congruence means that two shapes are exactly the same in size and shape, while similarity means that two shapes have the same shape but may differ in size. In other words, congruent shapes are always similar, but similar shapes may not always be congruent.

4. Can we prove congruence using only angles?

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5. Is there a shortcut or formula for proving congruence?

There is no one formula or shortcut for proving congruence. It depends on the given information and the shapes being compared. However, we can use properties and theorems, such as the Pythagorean theorem and the congruence postulates, to help us prove congruence in specific cases.

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