Dealing with Prime Numbers!

In the Gaussian integers, we can write $(a+i)(a-i)(b+i)(b-i) = (c+i)(c-i)$ . Each of the four factors on the left has prime norm ( $N(x+yi) = x^2+y^2$ ), so is a prime in ${\mathbb Z}[i]$ . So we can write $c+i$ as a product of two of the factors (times a unit) and $c-i$ as a product of the other two (times a unit). (Since the norm of $c+i$ is the product of the prime norms of $a+i$ and $b+i$ , there should be one factor with an $a$ in it and one with a $b$ in it.)

Well, $(a+i)(b+i) = (ab-1)+(a+b)i$ . This is not a unit multiple of $c+i$ or $c-i$ unless $ab = 2$ , which leads to solutions $(1,2,3)$ and $(2,1,3)$ . On the other hand, $(a-i)(b+i) = (ab+1)+(a-b)i$ , which is a unit multiple of $c+i$ or $c-i$ if and only if $a$ and $b$ are consecutive integers.

So wlog for now we can write $b = a+1$ , which implies $c = a^2+a+1$ . But the only way that both $a^2+1$ and $(a+1)^2+1$ are both prime is if $a = 1$ , since one of them is even and so it must be $2$ . So this leads back to the two solutions we already found. So those are the only ones. The answer is $\fbox{2}$ .

NB: It is not true that $a$ and $b$ must be consecutive integers in general, e.g. $(2,8,18)$ . The requirement that both factors on the left be prime is important.