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I have a question about complex numbers (a+ib) and (c+id) as solutions to Pell’s equation, with a to d natural non-zero numbers. $(a+ib)^2 –n (c+id)^2 =1$ implies that $2iab=n2icd$ and therefore $ab=ncd$. The real part is $(a^2-b^2) –n(c^2-d^2)=1 $

If n, c and d are primes then I cannot find a solution. a=n implies that b=cd, and the LHS becomes

$ n^2-(cd)^2 – n(c^2-d^2)= n^2-nc^2 + nd^2-(cd)^2 = n(n-c^2) +d^2(n-c^2)= (n+d^2)(n-c^2)=1$

which is impossible. b=n leads to a similar contradiction. Can non-primes factors or other ideas give a solution?

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  • $\begingroup$ I presume that you intend $a,b,c,d$ to be integers, not (nonnegative) natural numbers? $\endgroup$
    – Lubin
    Jun 17, 2013 at 15:25

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Your question almost exactly asks for a description of the units in the biquadratic field $K_n=\mathbb Q(i,\sqrt n\,)=\mathbb Q(\sqrt{\pm n}\,)$, just as solving Pell’s equation in the classical form asks precisely about the units in $\mathbb Q(\sqrt n\,)$. To be even more exact, in the classical case I’m talking about the invertible elements of $\mathbb Z[\sqrt n]$ if $n\equiv2,3\pmod4$ and of $\mathbb Z[(1+\sqrt n)/2]$ if $n\equiv1\pmod4$. The division into cases occurs because of the differing description of the ring of integers of $\mathbb Q(\sqrt n\,)$ in the two cases.

In any event, the solutions to Pell’s equation in the classical case differ from the statement about units only finitely, in the sense that the solutions form a subgroup of finite index in the group of all units. The same happens in your extended case, where you’re looking for solutions to Pell with coordinates in the Gaussian Integers. In your $K_n$, there is still a finite (cyclic) subgroup of roots of unity, and an infinite cyclic group, so that every unit is uniquely expressible as $\zeta u^m$, where $\zeta$ is a root of $1$, and the $u$ is a generating unit like $8+3\sqrt7$ in the case $n=7$ of the classical problem.

The upshot is that your solutions form a group that contains the group of classical solutions as a subgroup of finite index. That is, there may be a new solution, but some power of it will look like the original basic solution.

There are plenty of people who know much more about this particular question than I do, but I have time, and if you’d like to discuss it further, find my e-mail on my web page, and we can chat.

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