Extra solutions when multiplying by complex conjugate

Question

Where is the error in this reasoning?

Suppose I write $11$ as a product of Gaussian integers, $11=(a+bi)(c+di)$, and I want to determine for which $a,b,c,d$ the equation holds. Then multiplying both sides by the respective complex conjugate we get $121=(a^2+b^2)(c^2+d^2)$, which has a solution $a=0,b=11,c=1,d=0$, but this implies from the first equation that $11=11i$, so I'm assuming it's to do with multiplication by the units in $\mathbb Z[i]$, or similar to how squaring an equation can create more solutions, but could someone give a proper explanation?

Answer 1

$(a^2+b^2)(c^2+d^2)=121$ doesn't necessarily mean $(a+bi)(c+di)=11$. It could just as well mean $(a+bi)(c+di)=11i$, or $-11$, or $-11i$.

So there are several different possible values of $(a+bi)(c+di)$ that all give $(a^2+b^2)(c^2+d^2)=121$, and just from $(a^2+b^2)(c^2+d^2)=121$ you can't tell which one you're after.

Answer 2

Factorization into primes in $\Bbb Z[i]$ is unique up to order and units. Hence

1. Determine a factorization.

2. If you have $n$ prime factors, you can multiply each with a unit, resulting in $4^{n-1}$ possibilities. If you allow a leading unit in the representation, then it's $4^n$ possibilities because you effectively have one additional factor in the product.

3. In addition, you can rearrange the prime factors in $n!$ ways provided each factor occurs only once. If there are higher multiplicities, it's still simple combinatorics.

In the case of 11 there is only the prime factor 11. If you use a leading unit, then it's the 4 representations $u^{-1}\cdot(11u)$ for the 4 units $u$.