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 $\displaystyle 8q+1$ where $\displaystyle q $ is an integer

Answer

We wish to prove that $\displaystyle 8q+1=k^2$, where $\displaystyle k $ is an odd integer

Let $\displaystyle k $ be an odd integer of the form: $\displaystyle 2I+1$

$\displaystyle (2I+1)^2=4I^2+4I+1$

As $\displaystyle I $is an integer, it can either be odd or even

Case 1: $\displaystyle I $is odd, i.e. it is of the form:$\displaystyle 2m+1$

$\displaystyle (2I+1)^2= 4(2m+1)^2+4(2m+1)+1$

$\displaystyle 16m^2+16m+4+8m+4+1=16m^2+24m+9$

$\displaystyle (16m^2+24m+8)+1 $

$\displaystyle 8(2m^2+3m+1)+1$ Let $\displaystyle 2m^2+3m+1=q$

$\displaystyle 8q+1$

Case 2:$\displaystyle I $is even, i.e. it is of the form $\displaystyle 2m$

$\displaystyle (2I+1)^2= 4(2m)^2+4(2m)+1$

$\displaystyle 16m^2+8m+1=(16m^2+8m)+1$

$\displaystyle 8(2m^2+m)+1$ Let $\displaystyle 2m^2+m=q$

$\displaystyle 8q+1$

Proven in both cases, hence the statement is true.

End of solution. Q.E.D

I welcome your critiques and say thanks in advance,