So I need to prove that$\displaystyle n^2 -1 $ is divisible by 8 iff n is odd
I wrote n= 2P+1 where P = 2k
then $\displaystyle (2P+1)^2 - 1 = 4P^2+4P + 1-1 $ subing 2k back in I get
16k^2+8k = 8(2k^2+k)
Is this okay?
No. By setting $\displaystyle n=4k+1,$ you have only proven the statement for every other odd number: $\displaystyle \dots,\,-7,\,-3,\,1,\,5,\,9,\,13,\dots$.
Fortunately, you can fix it by handling the case for odd $\displaystyle P$ separately. Then you get $\displaystyle n=4k+3,$ and
$\displaystyle n^2-1=16k^2+24k+9-1=8(2k^2+3k+1).$
Hello, meg0529!
I need to prove that $\displaystyle X \:=\:n^2 -1 $ is divisible by 8 iff $\displaystyle n$ is odd
I wrote $\displaystyle n\:=\: 2p+1$, where $\displaystyle p = 2k$ . . . . no
$\displaystyle p$ does not have to be even.
$\displaystyle p$ can be any integer, and $\displaystyle n$ will be odd.
So we have: .$\displaystyle X \:=\:n^2-1 \:=\:(2p+1)^2-1 \:=\:4p^2 + 4p \:=\:4p(p+1)$
We see that $\displaystyle X$ is divisible by 4.
$\displaystyle \text{Note that: }\:X \:=\:4\cdot\!\!\!\!\!\!\!\!\underbrace{p(p+1)}_{\t ext{consecutive integers}} $
Since $\displaystyle p(p+1)$ is the product of two consecutive integers,
. . one of them is even, the other is odd.
Hence: .$\displaystyle p(p+1)$ is divisible by 2.
Therefore, $\displaystyle X$ is divisible by 8.
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It said if and only if, so we must prove it "the other way."
. . If $\displaystyle n^2-1$ is divisible by 8, then $\displaystyle n$ is odd.
We have: .$\displaystyle n^2-1$ is divisible by 8.
. . Then: .$\displaystyle n^2-1 \:=\:8k$, for some integer $\displaystyle k.$
And we have: .$\displaystyle n^2 \:=\:8k + 1$
We see that $\displaystyle 8k$ is even . . . hence: $\displaystyle 8k+1$ is odd.
. . So we have: .$\displaystyle n^2 \:=\:\text{odd}$
It can be shown that:
. . If the square of an integer is odd, the integer is odd.
The proof is quite simple, but omitted here.
Therefore: .$\displaystyle n$ is odd.
Certainly there is, and Soroban has provided it.
Take care.
Edit: Oh, and be sure you understand what is meant by "if and only if" (abbreviated iff). I misread your post initially and thought it only went one direction; there are basically two proofs that need to be made. Again, Soroban has this covered.
I see you have edited your post, but I will post my response anyway since it is already written.
No. An if-then statement means the implication goes in only one direction.
The statement "If $\displaystyle p,$ then $\displaystyle q,$" which can also be written as $\displaystyle \text{``}p$ only if $\displaystyle q,\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true (but it might be possible for $\displaystyle q$ to be true even though $\displaystyle p$ is false).
The statement $\displaystyle \text{``}p$ if $\displaystyle q,\text{''}$ means that if $\displaystyle q$ is true, then $\displaystyle p$ must also be true (but it might be possible for $\displaystyle p$ to be true even though $\displaystyle q$ is false).
The statement $\displaystyle \text{``}p,$ if and only if $\displaystyle q\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true, and if $\displaystyle q$ is true, then $\displaystyle p$ must also be true. Do you see how the implication goes both ways?
When proving a statement like $\displaystyle p\text{ if{f} }q,$ you need to prove both of these statements:
1. If $\displaystyle p,$ then $\displaystyle q$
2. If $\displaystyle q,$ then $\displaystyle p$