Question: How to factor $^{49}−$ over $GF(7)$ as a product of irreducible polynomials?
My approach: would like to know if it's okay, any help is appreciated.
$$\begin{align}P(x) &= x^{49} - x\\ &= x(x^{48} -1)\\ &= x(x^{24} +1)(x^{24} -1) \\ &= x(x^{12} - 1)(x^{24} +1)(x^{12}+1) \\ &= x(x^6-1)(x^6+1)(x^{24}+1)(x^{12}+1)\\ &=x(x^3-1)(x^3+1)(x^6+1)(x^{24}+1)(x^{12}+1)\\ &=x(x-1)(x+1)(x^2+1-x)(x^2+1+x)(x^2+1)(x^4+1-x^2)(x^8+1)(x^{16}+1-x^8)(x^4+1)(x^8+1-x^4)\\ &=x(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^2+1-x)(x^2+1+x)(x^4+1-x^2)(x^8+1-x^4)(x^{16}+1-x^8)\end{align}$$
For your convenience: (image below)
Another approach:
$GF(49)$ has dimension $2$ over $\mathbb F_7$. Over $GF(49)$ the polynomial $x^{49}-x$ has $7$ linear factors $x-t,t=0,...,6,$ and $21$ quadratic factors, one of which is $x^2+6x+6.$
Hence, working modulo $7$, if we set $x^2=x+1$, we can take $x+2$ as a generator (its first few powers are $x+2,5x+5,6x+1,5x+1,2x,6x+2,6x+3,5,...)$ because $(x+2)^t \neq 1$ for $1 \le t \le 47$ and, of course, $(x+2)^{48} = 1$. There are many other choices I guess.
Using $(\pm1)^2=1=-6$, $(\pm2)^2=4=-3$ and $(\pm3)^2=2=-5$ and using the completion of the square "from the outer term", $$x^4+ax^2+b^2=(x^2\pm b)^2+(a\mp2b)x^2,$$ you can further get (blue=incomplete factorization) \begin{align} (x^2+1−x)&=(x+3)^2-1=(x+2)(x-3)\\ (x^2+1+x)&=(x-3)^2-1=(x-2)(x+3)\\ (x^4+1)&=(x^2+1)^2-2x^2=(x^2+1)^2-(3x)^2=(x^2-3x+1)(x^2+3x+1)\\ (x^4+1−x^2)&=(x^2+3)^2-1=(x^2-3)(x^2+2)\\ \color{blue}{(x^8+1)}&=\color{blue}{(x^4-3x^2+1)(x^4+3x^2+1)}\\ (x^4-3x^2+1)&=(x^2-1)^2-x^2=(x^2-x-1)(x^2+x-1)\\ (x^4+3x^2+1)&=(x^2-1)^2+5x^2=(x^2-1)^2-(3x)^2=(x^2-3x-1)(x^2+3x-1)\\ \color{blue}{(x^8+1−x^4)}&=\color{blue}{(x^4-3)(x^4+2)}\\ (x^4-3)&=(x^2+2)^2-4x^2=(x^2-2x+2)(x^2+2x+2)\\ (x^4+2)&=(x^2-3)^2+6x^2=(x^2-x-3)(x^2+x-3)\\ \color{blue}{(x^{16}+1−x^8)}&=\color{blue}{(x^4-2x^2+2)(x^4+2x^2+2)(x^4-x^2-3)(x^4+x^2-3)}\\ (x^4-2x^2+2)&=(x^2+3)^2-x^2=(x^2-x+3)(x^2+x+3)\\ (x^4+2x^2+2)&=(x^2+3)^2-4x^2=(x^2-2x+3)(x^2+2x+3)\\ (x^4-x^2-3)&=(x^2-2)^2-4x^2=(x^2-2x-2)(x^2+2x-2)\\ (x^4+x^2-3)&=(x^2-2)^2-9x^2=(x^2-3x-2)(x^2+3x-2)\\ \end{align}