My teacher's solution:
a = x^4 - 1 ≡ 0 (5) ==> true for every x≡/≡0 mod (5)
b = x^4 - 1 ≡ 0 (11)==> x≡ +-1 mod (11)
And then solving the 4 systems to find the solutions
However my teacher doesn't explain anything, he just wrote this and I have no idea what he did and how he reached these solutions. I know how to do modulo x^2 but for x^4 I have no idea, or better I had the idea of doing t=x^2 and solving a quadratic modulo but it takes a lot and I'm pretty sure it's not the optimal or intended way
picture of the solution:
From $\; \begin{cases}x^4\equiv 1\mod 5\iff x\equiv \pm 1,\pm 2\mod 5,\enspace\\x^4\equiv 1\mod 11\iff x\equiv \pm 1 \mod 11,\end{cases} $ we deduce the solutions $\bmod 55$ thanks to the inverse isomorphism in the Chinese remainder theorem: let's start from a Bézout's relation between $5$ and $11$: $$1\cdot 11-2\cdot 5=1.$$ The inverse isomorphism is given by \begin{align} \mathbf Z/5\mathbf Z\times\mathbf Z/11\mathbf Z&\longrightarrow \mathbf Z/55\mathbf Z \\ (x\bmod 5,y\bmod11)&\longmapsto x\cdot1\cdot 11-y\cdot 2\cdot 5=(11x-10y)\bmod55. \end{align} You should find $8$ solutions: $$x\equiv \pm 1,\pm12,\pm 21\pm 23\mod 55.$$