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,
[itex]Z[/itex] is the set of all integers, ie: [itex]\mathbb{Z} = \{..., \ -3, \ , -2, \ , -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/itex]
Do you get it now?
Viet Dao,
[tex]n \in \mathbb{Z}[/tex] means that n is an integer.
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,
An odd integer: [tex]2n + 1, \ n \in \mathbb{Z}[/tex]
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,
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,