What we know is that a quaternion "prime" is defined to have a norm which is prime in Z otherwise the norm will end up being the product of two quaternions of smaller norm. I was originally thinking that we should adopt the proof of showing that there are infinitely many primes in Z (see page 2 of the book) by assuming a finite number of quternion primes q

_{1},q

_{2},...,q

_{k}and then forming a product of their norms and then adding 1. This new natural number is then not divisible by any of the norms norm (q

_{1}), norm (q

_{2}), ..., norm(q

_{k}). But this hits snags straight away because you don't know if this new number is prime in Z and you don't know if it even relates to a quaternion.

After much thought, here is my proof.

Let us suppose to the contrary that there are only a finite number of quaternion primes q

_{1},q

_{2},...,q

_{k}(where k is some natural number). Now the norm of each of these quaternion primes must be a prime in Z, so norm (q

_{1})=p

_{1}, norm (q

_{2})=p

_{2}, ... ,norm(q

_{k})=p

_{k}where p

_{1},p

_{2},...,p

_{k}are primes in Z.

Now there are an infinite number of primes in Z of the form 4n+1 (see page 113) and so we can always choose a prime p of this form which is not one of p

_{1},p

_{2},...,p

_{k}. Further by the two-square theorem (see page 109) p=a

^{2}+b

^{2}for some a, b in Z. Now this is the norm of quaternion q where q=a

**1**+b

**i**+0

**j**+0

**k**since det (q)=a

^{2}+b

^{2}but q is not one of q

_{1},q

_{2},...,q

_{k}since norm (q) is not any one of norm (q

_{1}), norm (q

_{2}), ..., norm (q

_{k}). Further q must be a quaternion prime since norm (q) is prime in Z. Therefore we have reached a contradiction. It follows that there must be infinitely many quaternion primes.