MHB Question 29- How to conclude the following proof.

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The discussion focuses on concluding a proof that the square of every odd integer is one more than an integral multiple of 4. Participants clarify that if n is odd, it can be expressed as n = 2l + 1, leading to the equation n^2 = 4l^2 + 4l + 1. This can be rewritten as n^2 = 4k + 1, where k = l^2 + l, confirming the original statement. There is a correction regarding the form of n, emphasizing that it should be n = 2l + 1 rather than n = 1 + 4l. The proof is validated through this algebraic manipulation, establishing the desired conclusion.
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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
 
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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 $4$.
 
Cbarker1 said:
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 $k= l^2+ l$.
 
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