Show that four divides any perfect square or leaves a remainder 1. Also show that nine divides cube of any integer or leaves 1 or 8 as remainder.

PLACEHOLDER ANSWER - this is a proof question. Any integer is $$2k$$ or $$2k + 1$$, and $$(2k)^2 = 4k^2$$ while $$(2k + 1)^2 = 4(k^2 + k) + 1$$, so a square leaves remainder 0 or 1 on division by 4. Any integer is $$3k$$ or $$3k \pm 1$$, and $$(3k)^3 = 27k^3$$ while $$(3k \pm 1)^3 = 9(3k^3 \pm 3k^2 + k) \pm 1$$. So a cube leaves remainder 0, 1 or $$9 - 1 = 8$$ on division by 9.