<s>[INST] Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers... Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers. Prove that $7$, $1 + 2\sqrt {5}i$, and $1 - 2\sqrt {5} i$ are all prime in $C$.
-I am really lost in this question, the closest looking thing I could muster up from our lecture notes was the section covering the fundamental theorem of arithmetic, and more specifically covering uniqueness. Really not sure how to approach this problem, any help and jump start is appreciated.
 [/INST] Hint: Use the norm: $N(a+b\sqrt 5i)= a^2+5b^2$.
Suppose $7=ab$. This implies $N(u)N(v) = N(7)=49$. 
$N(u)=1, N(v)=49$, write $u=a+i\sqrt5 b$, $N(u) =a^2+5b^2=1$ implies $b=0, a^2=1$.
$N(u)=7=N(v)$. Write $u=a+ i\sqrt5 b$, $N(u) =a^2+5b^2=7$ we deduce that $\mid b\mid \leq 1$. Suppose that $\mid b\mid =1$, $a^2 +5=7$, $a^2=2$ impossible.
Suppose that $b=0$, $a^2=7$ impossible.
Write $1+2i\sqrt 5 =uv$, $N(u)N(v) =21$. 
$N(u)=1, N(v)=21$ implies as above $u=1$.
$N(u)=3, N(v)=7$. The argument above shows you can't have $N(v)=7$.
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