Click here for part 1 containing propositions 1, 2 and 3.
Proposition 4
Claim: If a and b are relatively prime then every factor of a2+b2 is a sum of squares.
First, let a and b be a relatively prime pair; that is their greatest common divider (gcd) is 1.
Then let a2+b2=π0. π0 may or may not have the following factors:
Factor pairs of π0
Choosing a factor of π0, say x, we get that
Equation 1 a= xm±c,b= xn±d,with c, d≤12|x| and m, n∈R. So
Equation 2 a2+b2=m2x2±2mxc+c2+n2x2±2nxd+d2=Ax+(c2+d2),with A=m2x+2mc+n2x+2nd. Since x divides the left hand side of equation 2, x must also divide the right hand side and most importantly c2+d2, say c2+d2=yx.
The next part of the proof is split into two cases that depend on the choices of c and d.
c, d are not relatively prime (gcd≠1)
If c and d are not relatively prime, then their gcd is relatively prime to x. To prove this let α=gcd of c and d. Since α is a factor of both c and d they can be rewritten as:
Equation 3 c=αe,d=αf,for a relatively prime pair of positive integers e and f which must be relatively prime to ensure that c and d are relatively prime. Plugging these into equation 1 we have that
Equation 4 a= mx±αe,b= nx±αf,which shows that α (gcd of c and d) is relatively prime to x otherwise a and b would have a common factor which violates the condition that they are themselves relatively prime.
Additionally, we have that
Equation 5 c2+d2=α2e2+α2f2=α2(e2+f2),is equal to yx for some integer y. This implies that α2 divides y since it divides c2+d2 and it follows that
Equation 6 e2+f2=zx,for an integer z such that z=yα2 for a relatively prime e and f and z no more than half of x, since
Equation 7 zx=e2+f2≤c2+d2≤(x2)2+(x2)2=12x2,which follows since e and f are some factor of c and d and c, d≤12|x|.
- Now assume that x is not a sum of squares. Proposition 3 states that
If a number which can be written as a sum of two squares is divisible by a number which is not a sum of two squares, then the quotient has a factor which is not a sum of two squares.
Since zx is a sum of squares (from equation 7) and is divisible by x which is not a sum of squares, then from proposition 3 there must exist a factor, say w that divides z that is not a sum of squares.
This results in infinite descent; for a given x which is not a sum of squares there exists a smaller number w also not the sum of squares but both divide e2+f2, a sum of squares. Since the smallest integer factor of π0 is 1, there cannot exist a smaller number w that is divisible by a sum of squares since any non-zero sum of squares is at least 1, so x has to be a sum of squares.
Since x can be any factor of a2+b2 then all factors are a sum of squares, as proposed.
c, d are relatively prime
In this case, c and d can be used instead of e and f in equation equation 7 and the arguments that follow still hold true.
Example
Let a=19 and b=9. Then a2+b2=192+92=442.
Factor pairs of 442
It can be shown that all factors of a2+b2 are sum of squares:
Factors of 442 and their sum of squares
Proposition 5
Claim: Every prime of the form 4n+1 is a sum of two squares.
To say a prime number is in the form p=4n+1 is equivalent to saying p≡1 mod 4. Fermat’s Little Theorem guarantees that the numbers 14n, 24n, 34n, …, (4n)4n≡1 mod 4. Since all elements of the sequence are divisible by p, so are their differences:
p| 24n−14n,p| 34n−24n,p| 44n−34n,⋮p| (4n)4n−(4n−1)4n.Now each of these differences can be factorised as follows:
24n−14n, 34n−24n, ⋯=(22n+12n)(22n−12n), (32n+22n)(32n−22n), …So since p divides each of the differences it must also divide at least one of the pair of brackets (by Euclid’s Lemma):
Equation 1 a4n−b4n=(a2n+b2n)(a2n−b2n).If p divides all of the first brackets (with the sum of two squares) then Fermat’s Sum of Two Squares proof is complete; we have the sum of two relatively prime squares that is divisible by a prime p. Now proposition 4 states that:
If a and b are relatively prime, then every factor of a2+b2 is a sum of squares,
so p is a sum of squares since we are assuming it divides a sum of squares as originally proposed.
However, it could be the case that p divides the second pair of brackets. We will show that this is not possible by using the method of Finite Differences.
Assume p divides all of the first consecutive differences of the second brackets, that is p| 22n−1, 32n−22n, …, then p can also divide the 2nd differences: p| 32n−2(2)2n+1, 42n−2(3)2n+22n, … and in general, divide the kth difference. From section 3, it was shown that the kth difference of the sequence 1k, 2k, 3k, … is equal to k!. Substituting 2n instead of k would show that the 2nth difference would be equal to (2n!):
(4n)2n−(12n)(4n−1)2n+(22n)(4n−2)2n−(32n)(4n−3)2n+⋯=2n!,which is not divisible by p since p=4n+1>2n×2n−1×… 2×1, so p cannot divide the second bracket and is therefore, a sum of two squares.
Example
Take the first 4 prime numbers of the form 4n+1; 5, 13, 17, 29. It can be easily shown that they are themselves sum of squares:
4n+1 primes and their sums of squares
Although we consider Euler’s proof to be the most interesting, there are a handful of other proofs that use ideas from many different areas of Mathematics including Calculus and Topology.