I am having trouble understanding page 53, of A Classical Introduction to Modern Number Theory, by Kenneth Ireland and Michael Rosen.
Corollary 3. $$(-1)^{(p-1)/2} = \left(\frac{-1}{p}\right)$$ Where the right-hand side is the Legendre symbol.
The part I'm tripping over is the next paragraph, which states: Corollary 3 is interesting. Every odd integer has the form $4k + 1$ or $4k + 3$. (I understand that). Using this, one can restate the corollary as:
$$x^{2}\equiv -1\bmod(p)\text{ has a solution} \iff p \text{ is of the form } 4k + 1$$
How did the authors go from the corollary to the restatement?
Best Answer
What is the Legendre symbol? By definition $(\frac{-1}{p})=1$ if $(-1)$ is a quadratic residue mod $p$, otherwise $(\frac{-1}{p})=-1$. Now the statement says that $(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}$. Now it is easy to check that if $p$ has the form $4k+1$ then $\frac{p-1}{2}$ is even and hence by the statement $(\frac{-1}{p})=1$, which means $-1$ is a quadratic residue mod $p$. On the other hand if $p$ has the form $4k+3$ then $\frac{p-1}{2}$ is odd and hence $(\frac{-1}{p})=-1$, so in that case $-1$ is not a quadratic residue mod $p$.