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?

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