4K+1 Prime Unique Sum of Two Squares
Proof:
Consider: p=x**2 + y**2
Suppose p|x, then p|y => p**2|p, clearly false, so WLOG hcf (p,x)=1
Therefore x**2 + kp = -(y**2) has solution x0, y0 s.t.
x0**2 + y0**2 = kp
WLOG, take 0 < |x0| < p**(1/2), 0 < |y0| < p**(1/2)
=> x0**2 + y0**2 < 2p
i.e. x0**2 + y0**2 = p
Proof:
Suppose p=a**2 + b**2=c**2 + d**2
a**2d**2 - b**2c**2=p(d**2 - b**2)==0 [mod p]
ad==+/-bc [mod p]
Since a,b,c,d all <p**(1/2),
ad-bc=0 or ad+bc=p
Consider second equality:
p**2=(a**2 + b**2)(c**2 + d**2)=(ad + bc)**2 + (ac - bd)**2
=p**2 + (ac-bd)**2
=> ac-bd=0
So, either ad=bc or ac=bd
Suppose, for instance, that ad=bc. Then a|bc, with hcf (a,b)=1, which forces a|c, i.e. c=ka
Therefore ad=bc=b(ka), i.e. d=bk
Now, p=c**2 + d**2=k**2(a**2 + b**2) => k=1 => a=c, b=d
Similarly, if ac=bd, a=d, b=c