Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.

  
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*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
  

Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^2$, with 
(i) $(x + y\sqrt{-5}) = 1$, (ii) $(x - y\sqrt{-5}) = 1$
Adding both (i) & (ii), get: $x = 1$, Subtracting (ii) from (i), get: $(y\sqrt{-5}) = 0$.
Unable to pursue after that, as $x=-1$ is not possible.


  
*Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$
  

The hint given is to use the unique factorization theorem for the integers, by supposing $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$. The hint asks to show that: $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. I need some more hint or help to pursue, as squaring $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$ does not lead to $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. My attempt is stated for squaring both sides below:L.H.S.: $(x + y\sqrt{-5})(x + y\sqrt{-5}) => x^2 + 2xy(-5) -5y^2$  R.H.S.: $(a + b\sqrt{-5})^2(c + d\sqrt{-5})^2 => (a + b\sqrt{-5})(a + b\sqrt{-5})(c + d\sqrt{-5})(c + d\sqrt{-5}) => (a^2 -5b^2 +2ab\sqrt{-5})(c^2 -5d^2 +2cd\sqrt{-5})$


  
*Find all primes less than 50 in integers that can be written in the form $x^2 + 5y^2$.
  

No clue except to first find the primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.
Then trying to see if the factorization works, so starting with $2 = x^2 + 5y^2$, but cannot think further. Do I need to have $y$ as a imaginary number only, as $\sqrt{-5}$, or anything will work.
 A: *

*You're overthinking this one. We have two integers, $x$ and $y$, with $x^2 + 5y^2=1$. If $y$ is anything other than $0$, then $5y^2$ is greater than $1$, and since $x^2$ is non-negative, the equation is impossible. Having decided $y=0$, we are left with $x^2=1$, which has two solutions: $x=\pm 1$

*This question seems to be missing something. It is not true that every number of the form $x+y\sqrt{-5}$ is prime in $\Bbb Z[\sqrt{-5}]$. For example, we can take $x=1, y=5$, and note that $1+5\sqrt{-5}$ factors as $(3+\sqrt{-5})(2+\sqrt{-5})$.

*An easier way to do this is to pick values of $y$, and then for each one, try values of $x$ until you pass $50$.
A: Both $x$ and $y$ are drawn from $\mathbb Z$, right?


*

*If $x = \pm 1$, then $x^2 = 1$. Then $5y^2 = 0$, so obviously $y = 0$. Okay, I kinda took that backwards. Going forward: if $y \in \mathbb R$, then $y^2 \geq 0$. Therefore, if $y \neq 0$, then $5y^2 > 4 > 1$.

*Unless there is some restriction on $x$ and/or $y$, this statement is false. Simply choose $x$ and $y$ such that $\gcd(x, y) > 1$. Then $\gcd(x, y) \mid (x + y \sqrt{-5})$. Of course, as Tony showed, $\gcd(x, y) = 1$ is no guarantee either.

*Try fixing $y = 0$, then... um, that doesn't work, squares can't also be prime. Okay, try fixing $y = 1$, then $x^2 + 5y^2$ gives the sequence 5, 6, 9, 14, 21, 30, 41, from which you cull out 5 and 41. Then fix $y = 2$, giving 20, 21, 24, 29, 36, 45, which gives just 29. No need to bother with $y = 3$ for the specified range. So we've got 5, 29, 41. Just three primes, but that's enough to find http://oeis.org/A033205 in the OEIS.

A: For 3, the squares modulo $5$ are $-1,0,1$, so the primes are: $5,11,19,29,31,41$.
Question 2 doesn't make sense. Need provide context.
