Question 29- How to conclude the following proof.

[cbarker1]

Dear Everyone,
I have trouble writing the conclusion of the proof.
29. The square of every odd integer is one more than an integral multiple of 4.
Work:
Let $n\in\Bbb{Z}$
If n is odd, then $n^2=1+4k$ for some $k\in\Bbb{Z}$.

Examples
Let n=3. Then k=2.
Let n=5. Then k=6.
Let n=21. Then k=110.

Proof:
Suppose n is odd. Then $n=1+4l$ for some $l\in\Bbb{Z}$.
Then, $(2l+1)^2=1+4k$
$4l^2+4l+1=1+4k$
$4(l^2+l)+1=1+4k$
Since $4(l^2+l)\in\Bbb{Z}$. Then...

Thanks for the Help,
CBarker1

 

[Greg]

If $n$ is an odd integer then it is of the form $2k-1$. Squaring, we find $(2k-1)^2=4k^2-4k+1=4(k^2-k)+1$ hence $n^2$ is one more than an integral multiple of $4$.

 

[HallsofIvy]

Dear Everyone,
I have trouble writing the conclusion of the proof.
29. The square of every odd integer is one more than an integral multiple of 4.
Work:
Let $n\in\Bbb{Z}$
If n is odd, then $n^2=1+4k$ for some $k\in\Bbb{Z}$.

Examples
Let n=3. Then k=2.
Let n=5. Then k=6.
Let n=21. Then k=110.

Proof:
Suppose n is odd. Then $n=1+4l$ for some $l\in\Bbb{Z}$.

No. This is not true for n= 3 or 7 or 11, etc. What is true that $n= 1+ 2l$.

Then, $(2l+1)^2=1+4k$

Okay, so the "4" before was a typo. Now, this is what you want to prove.

$4l^2+4l+1=1+4k$
$4(l^2+l)+1=1+4k$
Since $4(l^2+l)\in\Bbb{Z}$. Then...

Thanks for the Help,
CBarker1

Not quite a valid proof because in stating "$(2l+ 1)^2= 4k+ 1$" you are assuming what you want to prove.

Instead, starting with n is odd, so $n= 2l+ 1$, we have $n^2= (2l+ 1)^2= 4l^2+ 4l+ 1= 4(l^2+ l)+ 1= 4k+ 1$ where $k= l^2+ l$.