Lorena Beard

2022-06-29

We can assume that $p\ne 2,3$ because 0 is a root of $f\left(x\right)$ in ${\mathbb{F}}_{2}$ and ${\mathbb{F}}_{3}$. Let Q be the "squares subgroup" (?) of index 2 in ${\mathbb{F}}_{p}^{\ast }$. If 2 and 3 are not in Q (this means that $\left({x}^{2}-2\right)\left({x}^{2}-3\right)$ doesn't have roots in ${\mathbb{F}}_{p}$), then $2\cdot 3\in Q$ (?), and then ${x}^{2}-6$ has a root in ${\mathbb{F}}_{p}$

Karla Hull

Expert

By the squares subgroup, they mean the subgroup

In other words, Q is the image of the homomorphism $\phi :{\mathbb{F}}_{p}^{\ast }\to {\mathbb{F}}_{p}^{\ast }$ given as $x↦{x}^{2}$
In the below, I am also going to assume that $p\notin \left\{2,3\right\}$
Since $\mathrm{ker}\phi =\left\{1,-1\right\}$ and $-1\ne 1$ (why?), we see that G has index 2 in ${\mathbb{F}}_{p}^{\ast }$
Now, if $2,3,6\notin Q$, then this means that the cosets $2Q,3Q,6Q$ are all not equal to Q. Since Q has index 2 in ${\mathbb{F}}_{p}$, this means that $2Q=3Q=6Q$
Thus, one of $2,3,6$ must be in Q. In turn, the given polynomial has a root in ${\mathbb{F}}_{p}$ (in fact, in ${\mathbb{F}}_{p}^{\ast }$, if $p\ne 2,3$)

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