# Math Help - Odd squares

1. ## Odd squares

Good night.
I have a proof question here that I needs validating. I was also wondering if I could receive alternative solutions for the question.
Thanks in advance for the help

Question
Prove that the square of an odd integer is always of the form $8q+1$ where $q$ is an integer

We wish to prove that $8q+1=k^2$, where $k$ is an odd integer
Let $k$ be an odd integer of the form: $2I+1$
$(2I+1)^2=4I^2+4I+1$

As $I$is an integer, it can either be odd or even
Case 1: $I$is odd, i.e. it is of the form: $2m+1$
$(2I+1)^2= 4(2m+1)^2+4(2m+1)+1$
$16m^2+16m+4+8m+4+1=16m^2+24m+9$
$(16m^2+24m+8)+1$
$8(2m^2+3m+1)+1$ Let $2m^2+3m+1=q$
$8q+1$

Case 2: $I$is even, i.e. it is of the form $2m$
$(2I+1)^2= 4(2m)^2+4(2m)+1$
$16m^2+8m+1=(16m^2+8m)+1$
$8(2m^2+m)+1$ Let $2m^2+m=q$
$8q+1$

Proven in both cases, hence the statement is true.
End of solution. Q.E.D

2. Hello, I-Think!

Prove that the square of an odd integer is always of the form $8q+1$ where $q$ is an integer.

Let the odd integer be: . $n \:=\:2p+1\,\text{ for some integer }p.$

. . Then: . $n^2 \:=\:(2p+1)^2 \:=\:4p^2 + 4p + 1$

Here is a very sneaky step . . .

$\text{We have: }\;n^2 \;=\;4\underbrace{p(p+1)} + 1\;\;{\color{blue}[1]}$
. . . . . . . .
two consecutive integers

With two consecutive integers, one of them is even.
. . Hence, their product is even: . $p(p+1) \:=\:2q\,\text{ for some integer }q.$

Then ${\color{blue}[1]}$ becomes: . $n^2 \;=\;4(2q) + 1 \;=\;8q+1 \quad\hdots\;\text{ta-}DAA!$