- #1

- 10

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Meh
- Start date

- #1

- 10

- 0

- #2

VietDao29

Homework Helper

- 1,424

- 3

An even integer can be written as [tex]2n, \ n \in \mathbb{Z}[/tex]

And an odd integer can be written as [tex]2n + 1, \ n \in \mathbb{Z}[/tex]

For example : 5 = 2 * 2 + 1 (n = 2), 7 = 2 * 3 + 1 (n = 3), 13 = 2 * 6 + 1 (n = 6).

So the quare of an add integer can be written as: [tex](2n + 1) ^ 2 = 4n ^ 2 + 4n + 1[/tex]

You will try to arrange [itex]4n ^ 2 + 4n + 1[/itex] into 8k + 1. Since you have '+ 1' in both sides, in fact, you just need to arrange [itex]4n ^ 2 + 4n[/itex] into 8k.

Can you go from here?

Viet Dao,

And an odd integer can be written as [tex]2n + 1, \ n \in \mathbb{Z}[/tex]

For example : 5 = 2 * 2 + 1 (n = 2), 7 = 2 * 3 + 1 (n = 3), 13 = 2 * 6 + 1 (n = 6).

So the quare of an add integer can be written as: [tex](2n + 1) ^ 2 = 4n ^ 2 + 4n + 1[/tex]

You will try to arrange [itex]4n ^ 2 + 4n + 1[/itex] into 8k + 1. Since you have '+ 1' in both sides, in fact, you just need to arrange [itex]4n ^ 2 + 4n[/itex] into 8k.

Can you go from here?

Viet Dao,

Last edited:

- #3

- 10

- 0

Hmm what does that "Z" in "n belongs to ..." mean? Not to fimilar with that sign :P

- #4

VietDao29

Homework Helper

- 1,424

- 3

Do you get it now?

Viet Dao,

- #5

- 10

- 0

Still don't get it >.< Sorry.

- #6

VietDao29

Homework Helper

- 1,424

- 3

And because:

[tex]\mathbb{Z} = \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/tex]

So:

[tex]n \in \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/tex]

That means n can take any value from that set, n can be -3, or -5, or 9, or 10, or 14, or 600, or -1004, ...

Viet Dao,

- #7

- 10

- 0

I understood that, just not the how to solve >.<

- #8

VietDao29

Homework Helper

- 1,424

- 3

So the square of an odd integer is: [tex](2n + 1) ^ 2 = 4n ^ 2 + 4n + 1[/tex].

You are going to prove [tex]\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n + 1 = 8k + 1[/tex]

+ 1 is in both sides, so you are going to prove:

[tex]\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n = 8k[/tex]

Note that:

[itex]4n ^ 2 + 4n = 4n(n + 1)[/itex]

What can you say about the product of two successive integers? ie : n * (n + 1).

Can you go from here?

Viet Dao,

- #9

- 10

- 0

Ah, alrite I can go from there. Thanks Viet Dao.

- #10

- 23

- 0

- #11

VietDao29

Homework Helper

- 1,424

- 3

k = 4:

8 * 4 + 1 = 32 + 1 = 33, and 33 is not a square of any integer.

k = 5:

8 * 5 + 1 = 40 + 1 = 41, and 41 is not a square of any integer.

Viet Dao,

Share: