This is a question in number theory. It seemed I can't start the proof.

Question

How can this be solved?

I knew that the concept of an order of an integer is needed, but rather than that I think I'm really lost with the proofs.

Answer 1

$$a^8-1\equiv 0 \implies (a^4+1)(a^4-1)\equiv 0 \implies a^4 \equiv -1 \pmod p$$

$$c^2=(a-a^3)^2=a^2-2a^4+a^6=a^2(1+a^4)-2a^4\equiv 2 \pmod p$$

$$d^2=(a+a^3)^2=a^2+2a^4+a^6=a^2(1+a^4)+2a^4\equiv -2 \pmod p$$

Answer 2

Hint: exploit innate symmetry: $\ \ \underbrace{\overbrace{-a^4 = 1^{\phantom{|}}_{\phantom{.}}\!\!}^{\!\!\!\!\textstyle \color{#0a0}{(-a^3)}a\!=\! 1}}_{\!\!\!\!\!\!\textstyle(\color{#c00}{-a^2})a^2\! =\! 1}\Rightarrow\ (a+\overbrace{a^{-1}}^{\!\!\textstyle \color{#0a0}{-a^3}})^2 = a^2+\underbrace{a^{-2}}_{\!\textstyle \color{#c00}{-a^2}}+ 2\, =\, 2$