RE: wonderings
05-27-2017, 08:38 AM
moreover we may exactly classify all numbers that can be written as a sum of two squares
lemma: Let p = 3 mod 4 and prime
if a^2+b^2 = 0 mod p then either a,b = 0 mod p (this is one solution) or we can divide by b (b is coprime to p so we can do this):
(a/b)^2=-1 mod p. So (a/b)^4=1 mod p, and thus the order of a/b in Z/pZ is 4. but wait, the cardinality of Z/pZ is p-1. and this is 2 mod 4. but by Lagrange's Theorem all element's order divide |Z/pZ| = p-1 so this is impossible!
therefore a,b = 0 mod p is the only solution.
(yeah ok we didn't really need group theory there, quadratic reciprocity does the trick too.)
anyway the punchline is that numbers with an odd number of any 3 mod 4 prime factor cannot be written as the sum of two squares. some minor details left out but you get the idea
(this is a necessary, and as it turns out, sufficient condition - what remains is to be able to show that all primes 1 mod 4 can be expressed as the sum of two squares, and you can use quadratic reciprocity)
lemma: Let p = 3 mod 4 and prime
if a^2+b^2 = 0 mod p then either a,b = 0 mod p (this is one solution) or we can divide by b (b is coprime to p so we can do this):
(a/b)^2=-1 mod p. So (a/b)^4=1 mod p, and thus the order of a/b in Z/pZ is 4. but wait, the cardinality of Z/pZ is p-1. and this is 2 mod 4. but by Lagrange's Theorem all element's order divide |Z/pZ| = p-1 so this is impossible!
therefore a,b = 0 mod p is the only solution.
(yeah ok we didn't really need group theory there, quadratic reciprocity does the trick too.)
anyway the punchline is that numbers with an odd number of any 3 mod 4 prime factor cannot be written as the sum of two squares. some minor details left out but you get the idea
(this is a necessary, and as it turns out, sufficient condition - what remains is to be able to show that all primes 1 mod 4 can be expressed as the sum of two squares, and you can use quadratic reciprocity)