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My question stems from the following: How many solutions $(x,y) \in \mathbb{N}$ are there to \begin{equation} x^2 + 6y^2 = 2^{100} \cdot 3^{100} \, ? \end{equation} This is the same as asking how many elements have norm $2^{100} \cdot 3^{100} $ in $\mathbb{Z}[\sqrt{-6}]$, however $\mathbb{Z}[\sqrt{-6}]$ is not a UFD. We can use Dedekind's Prime Factorisation Theorem to show the ideals of norm $2$ and $3$ are $\mathfrak{p}_2 = (2, \sqrt{-6})_R$ and $\mathfrak{p}_3 =(3, \sqrt{-6})_R$ respectively. I believe we then know, by unique factorisation of ideals, that the only ideal of norm $6^{100}$ is $\mathfrak{p}_2^{100} \mathfrak{p}_3^{100}$. How does one link this to the number of solutions? Is there only one solution, since there is only one ideal of this norm, and if so why is there this correspondence between ideals of norm $N$ and elements of norm $N$? Many thanks in advance for any help.

Tom
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It seems that there is an elementary argument. Suppose that $x^2+6y^2=6^n$ for some $n\ge 2$. Then $6\mid x$, so we can write $x=6r$. Then $6\mid y$ and we can write $y=6s$. So we obtain $$ r^2+6s^2=6^{n-2} $$ Now argue by descent.

In general, we can count the number of integer solutions as follows:

Counting the Number of Integral Solutions to $x^2+dy^2 = n$

Dietrich Burde
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  • Thanks for that pointer. It's useful to know there are easier tricks to find solutions in some of these cases. However I'm still interested in the number theory part of my question (it's from an algebraic number theory course, and the intention is to use ideals) so I will wait a short while before accepting your answer to see if anyone offers insight into the use of ideals when an elementary solution wouldn't be possible. – Tom Apr 22 '21 at 09:33
  • Your question is not only about finding a solution, but to count them ("how many solutions"). For this you'll need algebraic and analytic number theory in general - see the duplicate, and Dedekind's prime factorization is not needed. Perhaps the question was meant differently? Do you have a link to the question? – Dietrich Burde Apr 22 '21 at 09:39
  • The full question states: Find how many solutions there are for $(x, y) \in \mathbb{N}$ to $2x^2 + 3y^2 = 2^{99} \cdot 3^{100}$. Hints: Multiply by 2, work in $R = \mathbb{Z}[\sqrt{-6}]$, factorise $(\sqrt{-6})_R$. I'm thinking that the link is to finding principal ideals of norm $6^{100}$ of which I think there is 1, and therefore exactly 1 solution. In the same spirit, a later question asks for the number of solutions to $x^2 + 29y^2 = 2 \cdot 3^{51}$ and I got the answer 9 here by arguing that there are 9 principal ideals of norm $2 \cdot 3^{51}$ but I'm not sure if this is the right idea. – Tom Apr 22 '21 at 09:56
  • If there would be exactly one solution, you should find it. I don't see it. Is there perhaps a typo in $2x^2+3y^2=2^{99}\cdot 3^{100}$. Do you have a way to check this again? – Dietrich Burde Apr 22 '21 at 10:04
  • Yes, and I set another variable e.g. $z = 2x$ to start thinking about the question I posted - and then I could accept only the solutions for which $z$ is even to give the required $x$ – Tom Apr 22 '21 at 10:05
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    There is no solution $(x,y)\in \Bbb N$. Because then there would be one to $x^2+6y^2=6^0=1$ in positive integers, and that's impossible. And over the integers, only $x=\pm 1$, $y=0$, but $x=\pm 1$ is not even, contradiction. So a typo is still likely. – Dietrich Burde Apr 22 '21 at 10:08