Proving that there is no unique factorisation into S-primes Let $S = \{a+b*\surd(−5) : a, b ∈ \mathbb{Z}\}$. $S$ is a subset of $\mathbb{C}$. We can add and multiply elements of $S$ to get another element of $S$ and all the normal rules (as for $\mathbb{Z}$) hold. We can now define factorizations, divisibility, etc, in $S$ in the same way as for $\mathbb{Z}$. The units of $S$ are the elements $u$ such that there exists an element $v$ so
that $uv = 1$. A factorisation $t = xy$ in $S$ is called trivial if either $x$ or $y$ is a unit.
An element $t \in S$ is called an S-prime if it has no non-trivial factorisations. We define the norm of an element $s = a + b√(−5)$ to be $N(s) = a^2 + 5b^2(= ss^*)$.


*

*Show that there are no elements in $S$ of norm 2 or 3

*By considering factorisations of 6 and using norms, show that $S$ does not
have unique factorisation into S-primes


In (1) I can see that no $a$ or $b$ exist such that $a^2 + 5b^2 = 2$ and $a^2 + 5b^2=3$. However, I am not sure how to prove it more explicitly other than do this:
$a^2+5b^2=(a+b)^2 - 2ab + 4b^2$ has to be divisible by $2$ for the norm to exist. Clearly $2|2ab$ and $2|4b^2$ and so any linear combination will divide 2. But we then expect $2|(a+b)^2$ $\rightarrow$ $(a+b)|\surd(2)$ which is impossible since $a$ and $b$ are integers. Similarly, for the norm of $3$ we have that $3|(a+b)^2$ which is again not the case since $a$ and $b$ are integers.
In (2) I am not sure what approach to use. Following the hint I did that $N(6) = N(2*3) = N(2) * N(3)$
 But then if we consider $N(2)$ which implies that $t = a+b\surd(-5) = 2$ we have that there are multiple values of $a$ and $b$ that give $t$. For instance, $a=2$ and $b=0$ or $a=7$ and $b=-5$ and so on. I cannot see at the moment if this helps or not as I do not even know where to start.
I would appreciate any advise!
Thank you in advance!
 A: No, your proof for (1) is not correct. How did you "deduce" that "$(a+b)\mid\sqrt{2}$"? There are at least two issues with that. First, you can't switch the two numbers in a divisibility statement: $a\mid b$ does not imply that $b\mid a$ (so these two statements are not equivalent). Quick counterexample: $2\mid6$, but $6\nmid2$. Second, taking the square root of a divisibility statement (relation) is a highly questionable operation, exactly because it may introduce non-integers into a relation that was originally defined for integers only, so you end up with a meaningless statement.
But that proof can be done much easier. Let's say $a^2+5b^2=2$, where $a$ and $b$ are integers. If $b\neq0$, then $b^2\ge1$ and $a^2+5b^2\ge5$, so it can't be equal $2$. Therefore, $b$ has to be $0$. With $b=0$, we get $a^2=2$, which is impossible for integer $a$.
In the second part, you seem to be confused with inputs and outputs of the norm function. $N(2)$ does not mean the norm of a number being equal to $2$. On the contrary, here you're given a number, which is $s=2=2+0\sqrt{-5}$ (so $a=2$ and $b=0$), and then $N(2)$ is the norm of this number specifically, which is $N(2)=2^2+5\cdot0^2=4$.
The key to the second question is to find two different factorizations of $6$ into $S$-primes. You will have to prove that the numbers involved are indeed $S$-primes, and you will have to use part (1) for that. You already have one factorization $6=2\cdot3$. You need to show that both $2$ and $3$ are $S$-primes. For example, for $2$, you can start by observing that $N(2)=4$. If $2=pq$, a product of two non-units, then $N(p)N(q)=N(2)=4$, where neither norm can be $1$ (because they are non-units). Then the only remaining way to factor $4$ is $4=2\cdot2$, so $N(p)=2$ and $N(q)=2$ — but we already showed that this is impossible. Similarly for $3$.
This gives you one factorization of $6$ into $S$-primes. The other one is $6=\left(1+\sqrt{-5}\right)\left(1-\sqrt{-5}\right)$.
A: For 1, you can use the fact that $a^2$ and $5b^2$ are both positive. For there to be any chance of a solution, $b$ must be $0$ (otherwise $a^2+5b^2\ge 5$). It's then clear that no values of $a$ work.
For 2, the idea is to find a different factorisation of $6$ other than $2\times 3$. Norms come in because you still need to check that the elements in this factorisation are $S$-primes – there are primes of $\mathbb Z$ which are not primes in $S$.
For example, $2$ is irreducible. Indeed, if $ab = 2$, then $N(ab) = N(a)N(b) = 4$. This means that either 


*

*$N(a)$ or $N(b)$ is $1$, so $a$ or $b$ is a unit, or 

*$N(a) = 2$, which is impossible by part 1.

