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Meh

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- #1

Meh

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- #2

VietDao29

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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

Meh

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

- #4

VietDao29

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Do you get it now?

Viet Dao,

- #5

Meh

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Still don't get it >.< Sorry.

- #6

VietDao29

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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

Meh

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

- #8

VietDao29

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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

Meh

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

- #10

dracobook

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- #11

VietDao29

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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,

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