Unique Square Sum
You are given that ALL primes that are one more than a multiple of 4 can be written as the sum of two squares. For example, 13 = 22+32.
Assuming that a prime is expressible as the sum of two squares, prove that it can be done in only one way.
Suppose that the prime, p = a2+b2 = c2+d2.
Clearly HCF(a,b)=1, otherwise a2+b2 would be composite; similarly HCF(c,d)=1.
We will begin by establishing a result that will be used later.
If ad = bc then a|bc, but as HCF(a,b)=1, a|c. Let c = ka.
ad = kab d = kb.
So p = c2+d2 = (ka)2+(kb)2 = k2(a2+b2).
And because of the initial supposition, it is necessary that k = 1. Hence c = ka = a and d = kb = b.
By the symmetry of our initial supposition, from ad = bc we can freely interchange c and d to get ac = bd. And in the same way we can show that c = b and d = a.
Hence our proof will be complete if we can show that either ad = bc or ac = bd.
From the initial supposition, p = a2+b2 = c2+d2, we can write:
a2 = pb2 and d2 = pc2.
a2d2 = (pb2)(pc2) = p2p(b2+c2)+b2c2
a2d2 b2c2 mod p
ad bc mod p
So either (i) adbc 0 mod p, or (ii) ad+bc 0 mod p.
As 0 a2, b2, c2, d2 p
0 a, b, c, d p
0 ad, bc p
Hence -p adbc p and 0 ad+bc 2p.
If (i) is true then adbc 0 mod p, but as adbc lies between -p and p, it follows that adbc = 0 ad = bc. However, we have shown that if this is true, then c = a and d = b, and the square sum must be unique.
Let now consider (ii), ad+bc 0 mod p. We have shown that ad+bc lies between 0 and 2p, so if this is true then ad+bc = p. This case requires a useful, and tricky, identity to be obtained.
Hence if ad+bc = p, then p2 = p2+(acbd)2.
This leads to acbd = 0 ac = bd. However, we have shown that if this is true, then c = b and d = a, and once again we show that the square sum is unique.
Hence we prove that p can be written as the sum of two squares in only one way.