Euler factorization

Stream: maths

[Wouter Smeenk]

Is it possible to extend Euler factorization (https://en.m.wikipedia.org/wiki/Euler%27s_factorization_method#Theoretical_basis) to 3 factors? Meaning if n can be expressed as the sum of 2 squares in 4 ways it can be expressed as 3 factors. The converse follows from the Brahmagupta–Fibonacci identity applied twice. Is there some result on this? I could not find anything on this.

[Kevin Buzzard]

I don't quite understand the question but if a natural number can be written as the sum of two squares in four essentially different ways then this corresponds to four different factorisations in $\Z[i]$ which means that it can't be the product of <= 2 primes in $\Z$.

[Wouter Smeenk]

I mean if we have n = a^2 + b^2 = c^2 + d^2 = e^ + f^2 = g^2 + h^2 can we express n as n = (k ^ 2 + h ^ 2)(l ^ 2 + m ^ 2)(n ^ 2 + p ^ 2) in a way where we can calculate k,h,l,m,n and p using a,b,c,d,e,f,g and h (like is done for two factors in https://en.m.wikipedia.org/wiki/Euler's_factorization_method#Theoretical_basis)

[Kevin Buzzard]

The answer to that question is yes because (modulo the fact that you used h and n twice) you can just let k=l=0,h=m=1 and n=a,p=b.

[Kevin Buzzard]

(reordered)

[Kevin Buzzard]

The way to answer the question you probably mean to ask is "think about factorisations in Z[i]"

[Kevin Buzzard]

The idea is that n=a^2+b^2 can be reinterpreted as n=(a+bi)(a-bi) and Z[i] is a UFD.

[Wouter Smeenk]

Ah, I did not notice the duplicate use of n and h. That is indeed a solution, but I think there are also non trivial solutions. Is it possible to rewrite the proof in https://en.m.wikipedia.org/wiki/Euler's_factorization_method#Theoretical_basis using Z[i] and the reinterpretation you mentioned? I will look in to Z[i] more. Thanks for the help.