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

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