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Are there more classes of Diophantine equations which can be translated to unit elements in rings over $Z$? Basically I'm interested in classes of of nontrivial diophnatine equations with a parameter (no big deal if there isn't) like d in $x^2-dy^2=1$ that are proven to have infinitely many solutions for any $d$ (that satisfies a criterion like not being a square), and bonus points if solutions generate more.

The reason this interests me is that I've found those to be useful to prove existence of numbers that satisfy properties of the largest prime factor function. For example to find a number $a$ so that $f(a^2+1)<a$ where f is the largest prime factor function, I took the pell equations $a^2-2y^2=-1$ and deduced that there are infinitely many solutions.

Andy
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  • Certainly. Let $K$ be a number field and $\mathcal O_K$ be its ring of integers. This ring has a basis over $\mathbf Z$, say $e_1,\ldots, e_n$. The condition for $\alpha = \sum_{i=1}^n a_ie_i$ ($a_i \in \mathbf Z$) to be a unit in $\mathcal O_K$ is that the norm ${\text N}_{K/\mathbf Q}(\alpha)$ is $\pm 1$. The norm is a polynomial in the $a_i$'s. For example, if $K = \mathbf Q(\sqrt[3]{2})$ then $\mathcal O_K = \mathbf Z[\sqrt[3]{2}]$ and if we use the basis ${1,\sqrt[3]{2},\sqrt[3]{4}}$ then being a unit corresponds to solving $a^3 + 2b^3 + 4c^3 - 6abc = \pm 1$ for $a, b, c \in \mathbf Z$. – KCd Sep 24 '16 at 13:12
  • This link could be interesting for you: http://math.stackexchange.com/questions/1935613/fundamental-unit-of-the-quadratic-integer-ring-mathbbz-sqrtn/1935747#1935747 – Piquito Sep 24 '16 at 13:13
  • @KCd I know a bit about number fields, but not about it's ring of integers. Does what you say lead to diophantine equations with infinitely many solutions in pell-like but higher degree equations? – Andy Sep 24 '16 at 13:17
  • @Piquito Sorry not what I'm looking for. For example I want to ask if there are infinitely many solutions to $a^k+1=c*b^k$ for some $c,k$ – Andy Sep 24 '16 at 13:18
  • Every number field other than $\mathbf Q$ or an imaginary quadratic field has infinitely many units in its ring of integers (consequence of the Dirichlet unit theorem). If you want a parametric family, for non-cube $d \in \mathbf Z$ the units in $\mathbf Z[\sqrt[3]{d}]$ (which is not necessarily the full ring of integers of $\mathbf Q(\sqrt[3]{d})$ correspond to solutions $(a,b,c) \in \mathbf Z^3$ of $a^3 + db^3 + d^2c^3 - 3dabc = \pm 1$ for $a, b, c \in \mathbf Z$. – KCd Sep 24 '16 at 13:20
  • @KCd awesome, where can I read more about this? – Andy Sep 24 '16 at 13:20
  • If you want to look at the naive higher-degree version of Pell's equation, your plan is doomed: $x^3 - dy^3 = 1$ has at most one integer solution $(x,y)$ other than $(1,0)$. And there are similar results for $x^n - dy^n = 1$ for any $n > 2$. – KCd Sep 24 '16 at 13:21
  • Just read any book on algebraic number theory. – KCd Sep 24 '16 at 13:21
  • Try: http://www.springer.com/cda/content/document/cda_downloaddocument/9780387955292-c7.pdf?SGWID=0-0-45-101863-p2290488 – O. S. Dawg Sep 28 '16 at 04:05

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