x odd then exists y such that $x^2 = 8y+1$

Question

How can I algebraically show that if $x$ is odd, then $x^2 = 8y + 1$?

Making sure it is a true statement, I tabulated some values of $x,y$ pairs
$(1,0)$, $(3,1)$, $(5,3)$, $(7,6)$, $(9,10)$

I let $x = 2w + 1$ (since $x$ is odd)

Then $(2w + 1)^2 = 4w^2 + 4w + 1$

I don't think I can say $4w^2 + 4w + 1 \quad ? \quad 4w + 4w + 1 = 8w + 1$, can I?

Answer 1

If $x$ is odd, it is congruent to $\pm 1$ or $\pm 3$ modulo 8. In both cases, its square is congruent to $1$.

Answer 2

Because $$w^2+w=w(w+1)$$ is divided by $2$,

Answer 3

If $x = 8 k + 1,$ what happens? What if $x=8k + 3?$ You can do the rest.