Why does an even $x$ imply $y^2=-2 \pmod 8$

Question

I am very new to modular arithmetic, and I encountered the following statement on page 7 of this paper: If $x$ is even then $y^2 \equiv-2\pmod{8}$

The equation in question is $y^2=x^3-2$

I do not understand why. Could someone give a hint or a nudge in the right direction as to why this is true? My knowledge of modular arithmetic is sparse and I have absolutely no idea where to look in this case.

Answer 1

Write $x$ as a general even number, put $x=2n$ for $n \in \mathbb{Z}$, then $$y^{2}=(2n)^{3}-2=8n^{3}-2$$ Modulo $8$ this leaves a remainder of $-2$.

Answer 2

Consider the following cases:

• $x\equiv\color\red0\pmod8 \implies x^3-2 \equiv\color\red0^3-2\equiv 0-2\equiv 0\cdot8-2\equiv-2\pmod8$
• $x\equiv\color\red2\pmod8 \implies x^3-2 \equiv\color\red2^3-2\equiv 8-2\equiv 1\cdot8-2\equiv-2\pmod8$
• $x\equiv\color\red4\pmod8 \implies x^3-2 \equiv\color\red4^3-2\equiv 64-2\equiv 8\cdot8-2\equiv-2\pmod8$
• $x\equiv\color\red6\pmod8 \implies x^3-2 \equiv\color\red6^3-2\equiv216-2\equiv27\cdot8-2\equiv-2\pmod8$