Rotman wants me to prove the above in a very specific way:
Step 1: Prove that if p is a prime with $p\equiv 3$ mod 4, then either $a^2 \equiv 2$ mod p is solvable or $a^2 \equiv -2$ mod p is solvable. The hint says to note that the multiplicative group $F_p^\times$ is isomorphic to a group of order 2 times a group of odd order m, and the group of order m must contain 2 or -2 because $F_2 \times Z_m = (\{1\}\times H)\cup(\{-1\}\times H)$. Finally, note that every element in a group of odd order has a square root.
Pretty much all of that makes sense to me, except this: I believe we have to show the isomorphic copy of either 2 or -2 has to belong to the left part of the union, because in that case the isomorphic copy of the square root is just (1, square root of some element of H). But how do you do this? What prevents both 2 and -2 from mapping to the right part of the union?
Step 2: Prove that $x^4+1$ factors in $F_p[x]$, where p is an odd prime, if any of the following congruences are solvable:
- $b^2\equiv -1$ mod p
- $a^2 \equiv 2$ mod p
- $a^2 \equiv -2$ mod p
This was easy. I don't need help on step 2.
Step 3: Show $x^4 + 1$ factors in $F_p[x]$ for all primes $p$
It's obviously true for p=2. I also see why it's true for $p\equiv 3$ mod 4. But I don't see why it's true for $p\equiv 1$ mod 4. Maybe it's true that $p\equiv 1$ mod 4 implies $b^2\equiv -1$ mod p is solvable, which would finish the problem, but I don't know how to show that.
