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The Gaussian $\mathbb{Z}[i]$ and Eisenstein $\mathbb{Z}[\omega]$ integers have been used to solve some diophantine equations. I have never seen any examples of the golden integers $\mathbb{Z}[\varphi]$ used in number theory though. If anyone happens to know some equations we can apply this in and how it's done I would greatly appreciate it!

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    $\begingroup$ one place you might check are the archives of fibonacci quarterly. $\endgroup$ Jan 23, 2011 at 2:08
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    $\begingroup$ Not a solution to any diophantine equations, but something related to the Golden Ratio: jstor.org/stable/3029218. $\endgroup$ Jan 23, 2011 at 2:10
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    $\begingroup$ @Willie: In fact, according to Wikipedia, $\mathbb{Z}[\varphi]$ does have class number $1$. $\endgroup$ Jan 23, 2011 at 2:19
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    $\begingroup$ @quanta: Well, $\mathbb{Z}[\varphi]$ is a UFD in which you can decompose $x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy)$, giving one way to prove FLT for exponent 5. $\endgroup$ Jan 23, 2011 at 4:12
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    $\begingroup$ The interesting paper of Bergman "A number system with an irrational base" proposed by Arturo may be seen here. $\endgroup$ Jun 20, 2012 at 14:53

5 Answers 5

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Using the fact that $\mathbb{Z}[\varphi]$ is a unique factorization domain in which we can decompose $$ x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy),\quad\qquad{\rm(1)} $$ we can give a proof of Fermat's Last Theorem for the case of exponent 5. Here, I am using a bar over a number to denote conjugation, so $\varphi=(1+\sqrt{5})/2$ and $\bar\varphi=(1-\sqrt{5})/2$.

Theorem: There are no solutions to $x^5+y^5=z^5$ for nonzero $x,y,z$ in $\mathbb{Z}[\varphi]$.

That is, for exponent 5, FLT holds in the ring $\mathbb{Z}[\varphi]$ and, in particular, it holds in the integers.

Before going any further, let's note a few facts about factorization in $\mathbb{Z}[\varphi]$. As is well known, it is norm-Euclidean, so is a unique factorization domain. We have the prime factorizations $5=(\sqrt{5})^2$ and $11=q\bar q$, where I am setting $q=4-\sqrt{5}$ (for the remainder of this post). The identity $\varphi\bar\varphi=-1$ shows that $\varphi$ is a unit. In fact, it is a fundamental unit, so that every unit in $\mathbb{Z}[\varphi]$ is of the form $\pm\varphi^r$ for integer $r$. It will also be useful to use mod-q arithmetic (with $q$ as above). Then, $\varphi=(1+\sqrt{5})/2=8$ (mod q). Therefore every element of the quotent $\mathbb{Z}[\varphi]/(q)$ is equal to a rational integer mod q. As 11 = 0 mod q, this gives $\mathbb{Z}[\varphi]/(q)\cong\mathbb{Z}/(11)$. So, mod-q arithmetic in $\mathbb{Z}[\varphi]$ is exactly the same as mod-11 arithmetic in the integers. In particular, every 5'th power is equal to one of $0,1,-1$ mod q. Applying this to the equation $x^5+y^5=z^5$ shows that at least one of $x,y,z$ must have a factor of q. By dividing through by their highest common factor, we reduce to the case where $x,y,z$ are coprime, so exactly one is a multiple of q. Rearranging as $(-z)^5+y^5=(-x)^5$ if necessary, we can always bring the multiple of q to the right hand side. This reduces the problem to the following.

Theorem 2: There are no solutions to $x^5+y^5=uz^5$ for nonzero coprime $x,y,z\in\mathbb{Z}[\varphi]$ with $u\in\mathbb{Z}[\varphi]$ a unit and $q$ dividing $z$.

Let's prove this by showing that, if we have one solution, then we can find another solution for which $xyz$ has strictly fewer distinct prime factors. Applied to a minimal solution, this would give a contradiction. This is essentially the method of descent used by Fermat himself for the case of exponent 4.

So, suppose we have one solution. Writing $c_0=x+y$, $c_1=x^2+y^2-\varphi xy$ and $c_2=x^2+y^2-\bar\varphi xy$, (1) gives the decomposition $uz^5=c_0c_1c_2$. Also, $$ c_0^2-\bar\varphi c_1-\varphi c_2=0.\qquad\qquad{\rm(2)} $$ We would like to show that the factors $c_0,c_1,c_2$ are 5'th powers, which will be easier if they are coprime. Using the fact that $x,y$ are coprime to $z$, the identities $$ \begin{align} &c_0^2-c_1=\sqrt{5}\varphi xy,\\ &c_0^2-c_2=-\sqrt{5}\bar\varphi xy,\\ &c_1-c_2=-\sqrt{5}xy \end{align} $$ show that the highest common factor of $c_0^2,c_1,c_2$ is either 1 or $\sqrt{5}$. Consider the case where $\sqrt{5}$ divides $z$. Then it will also divide at least one of $c_i$, and the identities above show that it divides each $c_i$. In particular, 5 divides $c_0^2$, so the identities above show that $\sqrt{5}$ divides each of $c_1,c_2$ exactly once.

In the case where $z$ is not a multiple of $\sqrt{5}$, let us set $\tilde c_0=c_0^2,\tilde c_1=c_1,\tilde c_2=c_2$ and, in the case where $\sqrt{5}$ divides $z$, set $\tilde c_0=c_0^2/\sqrt{5},\tilde c_1=c_1/\sqrt{5},\tilde c_2=c_2/\sqrt{5}$. These are coprime and $$ \tilde c_0\tilde c_1^2\tilde c_2^2 = u^2\left(z^{2}/\sqrt{5}^{m}\right)^5 $$ where $m=0$ if $\sqrt{5}$ does not divide $z$ and $m=1$ if it does. As they are coprime, each prime factor of $z$ divides exactly one of the $\tilde c_i$, and its exponent is a multiple of 5. So, considering prime factorizations, each $\tilde c_i$ is equal to a unit multiplied by a fifth power $w_i^5$. So, (2) gives $$ u_0w_0^5+u_1w_1^5+u_2w_2^5=0 $$ for units $u_i$. Without loss of generality, we assume that $q$ divides $w_0$ and, dividing through by $-u_1$ if necessary, we suppose that $u_1=-1$. Then, $u_2=\pm1$ mod q. However, being a unit, we have $u_2=\pm(\varphi)^r=\pm 8^r$ (mod q), and, looking at this mod 11, 5 must divide $r$. So, $u_2$ is a fifth power and, by absorbing $-u_2^{1/5}$ into $w_2$, we can take $u_2=-1$. So we have arrived at $$ w_1^5+w_2^5=u_0w_0^5. $$ Also, all prime factors of $w_0w_1w_2$ are factors of $z$. So, except in the case where $x,y$ are units, we have a solution with strictly fewer prime factors, and we are done.

So suppose that we have a solution to Theorem 2. Iteratively applying the procedure above will keep generating new solutions and, as the number of prime factors of $xyz$ cannot decrease indefinitely, we must eventually settle on the case where $x,y$ are units, so that $x/y=\pm\varphi^r$. Exchanging $x,y$ if necessary, we suppose that $r > 0$. Then, $q^5$ is a factor of $1\pm\varphi^{5r}$. Using the identity $\varphi^5=-1+\varphi^4q$, and applying the binomial identity, it can be seen that $rq$ must be a multiple of $q^{5}$, so $r$ is a multiple of $11^4$. In particular, $\vert x/y\vert=\vert\varphi\vert^r$ will be very large (note, $\vert\varphi\vert^{11^4} > 10^{3000}$). Then, the definitions above for $c_0^2,c_1,c_2$ are dominated by the $x^2$ terms, so the ratios $\tilde c_i/\tilde c_j$ are close to one. Going through these details bounds the ratios $w_i/w_j$ and, in particular, none of them will be as large as $\vert\varphi\vert^{11^4}$. This means that we cannot have $x,y$ and $w_1,w_2$ all units. So, continuing the induction will generate solutions with ever fewer prime factors, giving the required contradiction.


This method of approaching FLT for exponent 5 was something I came up with after seeing the exponent 3 case in lectures years ago. It is a bit tedious having to separately deal with the case where $x,y$ are units. Maybe that can be tidied up. Essentially, the reason why this method works is because $\mathbb{Z}[\varphi]$ consists precisely of the real algebraic integers of the cyclotomic field $\mathbb{Q}(\zeta_5)$.

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  • $\begingroup$ Thanks for this! Would anyone be able to explain how the highest common factor of c_0^2,c_1,c_2 is found to be 1 or sqrt(5)? $\endgroup$
    – quanta
    Jan 24, 2011 at 0:32
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    $\begingroup$ Because it divides $\sqrt{5}xy$. And, as it divides a power of $z$ (so is coprime to $x,y$), it must divide $\sqrt{5}$. As this is prime, $\sqrt{5}$ and 1 are the only possibilities. $\endgroup$ Jan 24, 2011 at 0:36
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    $\begingroup$ btw, you can prove FLT for exponents 3,4,7 in a very similar way. For exponent 3, use $x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y)$ in $\mathbb{Z}[\omega]$. For 4, which is the easiest, use $x^4-y^4=(x-y)(x+y)(x^2+y^2)$ in the integers. For 7, which gets rather messy, use $x^7+y^7=(x+y)(x^3+y^3-\alpha x^2y-\bar\alpha xy^2)(x^3+y^3-\bar\alpha x^2y-\alpha xy^2)$ in $\mathbb{Z}[\alpha]$ with $\alpha=(1+\sqrt{-7})/2$. $\endgroup$ Jan 24, 2011 at 1:08
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    $\begingroup$ @George Lowther: +1, Very nice post. $\endgroup$ Jan 24, 2011 at 2:28
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    $\begingroup$ Because $w_0=0$, $u_1=-1$, $w_1^5=\pm1$ and $w_2^5=\pm1$ mod $q$. As I mentioned earlier up, mod $q$ arithmetic is the same as mod 11 in the integeres, so 5th powers are equal to 0,1 or -1 mod $q$. $\endgroup$ Feb 13, 2011 at 3:50
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While not nearly as impressive as George's amazing answer, it's worth noting that $\mathbb{Q}[\phi]$ (though not quite $\mathbb{Z}[\phi]$, the elements are in $\frac{1}{2}\mathbb{Z}[\phi]$) shows up in the icosians (a subgroup of order 120 of the group of unit quaternions) and the theory of the icosahedral group and the 600-cell (and even, tangentially, $E_8$); check out the Wikipedia page on the icosians for more details. (It's, loosely, a higher-dimensional version of the description of the vertices of the icosahedron as the corners of the golden rectangle $(0, \pm 1, \pm \phi)$ and the other two rectangles given by cyclic permutations of the coordinates)

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    $\begingroup$ +1 The tangential relation to $E_8$ and such comes from the fact that when we build the division algebra of the usual Hamiltonian quaternions, but with center restricted to $\mathbb{Q}(\phi)$ instead of all the reals, we get the algebra known as Icosians. Its non-trivial Hasse invariants occur only at the infinite places. Hence the maximal orders of this algebra have a very small discriminant, and therefore they are destined to give rise to dense lattices. $\endgroup$ Jun 20, 2012 at 11:11
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Formally the Riemann zeta-function can be expressed as

$$ \zeta(z)=\prod_{k=0}^{\infty}\;\;\prod_{p\in \mathbb{P}}\bigg\{\left( 1 -\varphi^{-1}\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\left( 1 +\varphi\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\bigg\} \;for\;z>1$$

where $ \varphi=\frac{1+\sqrt{5}}{2} $ is the Golden Ratio. This follows from the fact that the zeta function can be expressed as

$$ \zeta(z)=\prod_{p\in \mathbb{P}}\;\;\sum_{k=0}^{\infty}\frac{1}{p^{k\;z}}$$

and after some manipulations its easy to get the above representation.


To justify this, take the series $$ f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}+\cdots $$ we can write this as $$ f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}(1+x+x^{2}+x^{3}+x^{4})+x^{10}(1+x+x^{2}+x^{3}+x^{4})+\cdots $$ or equivalently $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}\cdots) $$ and again $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+\cdots) $$ or $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20})(1+x^{25}+x^{50}+x^{75}+x^{100}+\cdots) $$
so the general pattern is $$ f(x)=\prod_{k=0}^{\infty}(1+x^{1\cdot 5^{k}}+x^{2\cdot 5^{k}}+x^{3\cdot 5^{k}}+x^{4\cdot 5^{k}}) $$ now make $y=x^{5^{k}}$, one has that $$ 1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{\sqrt{5}-1}{2}y+1\right)\left(y^{2}+\frac{\sqrt{5}+1}{2}y+1\right) $$ where $\varphi=\frac{\sqrt{5}+1}{2}$ and $\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2}$ so $$ 1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{1}{\varphi}y+1\right)\left(y^{2}+\varphi y+1\right) $$ now remember that $$ \zeta(s)=\prod_{p \in \mathbb{P}}\left(1+\frac{1}{p^{s}}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}+\cdots \right) $$

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  • $\begingroup$ Could you please give me a reference to this formula? $\endgroup$
    – dot dot
    Jul 23, 2012 at 17:35
  • $\begingroup$ Where is it valid ? I assume Re (s) > 1. $\endgroup$
    – mick
    Sep 10, 2012 at 15:46
  • $\begingroup$ Oh i read 2,5 instead of 2.5 haha. Makes sense now. Thanks for the clarification. One more comment : i think it might be intresting to generalize this so that it holds for $Re(z)>0$. Who knows you might be able to solve the RH then ;) $\endgroup$
    – mick
    Sep 18, 2012 at 18:32
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You would probably solve the Mordell equation $y^2=x^3+5$ by working in that field.

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I wonder why Binet's formula to calculate Fibonacci number wasn't already posted. Here it is: $$ F\left(n\right) = {{\varphi^n-(-\varphi)^{-n}} \over {\sqrt 5}}. $$ Since you asked for $\mathbb Z(\varphi)$, reformulate it to $$ 5 F\left(n\right)^2 +2 = \varphi^{2n}+\varphi^{-2n}. $$ to get a nice identity for your Fibonacci integers.

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