- #1
MathematicalPhysicist
Gold Member
- 4,699
- 372
I forgot why the next statement is true and it's bugging me endlessly...
If p is prime such that p =1 mod 4 then (-1) = x^2 mod p.
Now in Ashe's Algebriac Number theory notes (book?), he says that
[tex]((\frac{p-1}{2})!)^2= -1 mod p[/tex]
I am quite stumped as how to show this, he argues we just need to look at:
1*2*...((p-1)/2)*(-1)*(-2)...*(-(p-1)/2)
which on the one hand because p =1 mod 4 it equals [tex]((\frac{p-1}{2})!)^2 [/tex]
on the other hand it also equals (p-1)!, but why does this equal -1?
Thanks.
If p is prime such that p =1 mod 4 then (-1) = x^2 mod p.
Now in Ashe's Algebriac Number theory notes (book?), he says that
[tex]((\frac{p-1}{2})!)^2= -1 mod p[/tex]
I am quite stumped as how to show this, he argues we just need to look at:
1*2*...((p-1)/2)*(-1)*(-2)...*(-(p-1)/2)
which on the one hand because p =1 mod 4 it equals [tex]((\frac{p-1}{2})!)^2 [/tex]
on the other hand it also equals (p-1)!, but why does this equal -1?
Thanks.