# Math - Proving

1. Sep 9, 2005

### Meh

Prove that the square of an odd integer is always of the 8k + 1, where k is an integer. Any help would be appreciated.

2. Sep 9, 2005

### VietDao29

An even integer can be written as $$2n, \ n \in \mathbb{Z}$$
And an odd integer can be written as $$2n + 1, \ n \in \mathbb{Z}$$
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: $$(2n + 1) ^ 2 = 4n ^ 2 + 4n + 1$$
You will try to arrange $4n ^ 2 + 4n + 1$ into 8k + 1. Since you have '+ 1' in both sides, in fact, you just need to arrange $4n ^ 2 + 4n$ into 8k.
Can you go from here?
Viet Dao,

Last edited: Sep 9, 2005
3. Sep 9, 2005

### Meh

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

4. Sep 9, 2005

### VietDao29

$Z$ is the set of all integers, ie: $\mathbb{Z} = \{..., \ -3, \ , -2, \ , -1, \ 0, \ 1, \ 2, \ 3, \ ... \}$
Do you get it now?
Viet Dao,

5. Sep 9, 2005

### Meh

Still don't get it >.< Sorry.

6. Sep 9, 2005

### VietDao29

$$n \in \mathbb{Z}$$ means that n is an integer.
And because:
$$\mathbb{Z} = \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}$$
So:
$$n \in \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}$$
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. Sep 9, 2005

### Meh

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

8. Sep 9, 2005

### VietDao29

An odd integer: $$2n + 1, \ n \in \mathbb{Z}$$
So the square of an odd integer is: $$(2n + 1) ^ 2 = 4n ^ 2 + 4n + 1$$.
You are going to prove $$\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n + 1 = 8k + 1$$
+ 1 is in both sides, so you are going to prove:
$$\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n = 8k$$
Note that:
$4n ^ 2 + 4n = 4n(n + 1)$
What can you say about the product of two successive integers? ie : n * (n + 1).
Can you go from here?
Viet Dao,

9. Sep 10, 2005

### Meh

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

10. Sep 10, 2005

### dracobook

hi, I'm wondering, isn't $$\forall k$$? cause k is stated to be an integer? just making sure..

11. Sep 10, 2005

### VietDao29

It can't be for all k, because:
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,