- #1

cbarker1

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MHB

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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