Golden Numbers

loopspace

25th February 2014

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Contents

  1. Home

  2. 1. Introduction

  3. 2. Golden Numbers

  4. 3. Relating the Golden Numbers

  5. 4. General Relations

  6. 5. The Golden Ratio

  7. 6. Golden Primes

  8. 7. Golden Conclusion

1 Introduction

This is a follow-up to my post on Euler's Identity. After complaining that I think Euler's identity is pretty ugly, I made the claim that the following equation was more beautiful.

2cos(ilog(1+ϕ))=3.

Actually, the original was 2sin(ilog(ϕ))=i but I prefer the above.

The occurrence of two similar but distinct equations involving ϕ led to me, and one or two others, to wonder about more such numbers. I had a bit of fun playing with the resulting identities and would like to record what I found out.

This is not a deep mathematical investigation. The paths it led me down are far from my own mathematical specialisation so I cannot be sure that I am even asking the right questions, let alone knowing which such questions have obvious or well-known answers and which don't.

The point is to show that even from something silly, it is possible just to have fun with mathematics.

2 Golden Numbers

Let's restate the equations:

2sin(ilog(ϕ))=i,2cos(ilog(1+ϕ))=3.

My goal was to start from eit=cos(t)+isin(t) and get something other than Euler's identity. Rearranging, we get the formulae for cos(t) and sin(t) in terms of eit and e-it. The breakthrough was the realisation that if x=eit then e-it=x-1 so I was looking at expressions of the form x+x-1 and x-x-1; to wit, the golden ratio. At that point, it is simply a matter of plugging them in and seeing which combinations give nice numbers.

But why stop there? The great thing about ϕ and 1+ϕ was that they solve, respectively, ϕ-ϕ-1=1 and (1+ϕ)+(1+ϕ)-1=3. The nice property here is that the right hand side consists of integers. But 1 and 3 are not the only integers. Let's define:

Definition 1

The Golden Numbers are the solutions of:

x±x-1

Specifically, for n let ϕn,1 be a solution of x-x-1=n and ϕn,-1 a solution of x+x-1=n.

To link back to the original purpose, we have:

2cosilogϕn,-1=n2sinilogϕn,1=ni

Notice that we say "a solution" instead of "the solution". There will usually be two, but knowing one tells us the other (via reciprocation and, for ϕn,1, a sign flip) and it is more convenient not to have to distinguish. Thus when we say "x=ϕn,α" we shall actually mean that x satisfies the equation for ϕn,α. The apparent sign flip in the definition is because it gives us the original golden ratio as ϕ1,1. We can also arrange the defining equation to be such that ϕn,α is a solution of:

x2=nx+α

which makes the sign convention a little more plausible. From this form, we can derive the formulae:

ϕn,α=n±n2+4α2

Of more interest, though, are relations between them.

3 Relating the Golden Numbers

The first observation to make is that ϕ-n,α=-ϕn,α. Thus we only need to consider n. The second is that ϕn,-1 is only real for |n|2, and ϕ2,-1 is unique in being single valued (i.e., 1).

Of more interest is finding relations between them corresponding to the original two equations involving the golden ratio itself. We have ϕ1,1=ϕ and so the original equations can be summed up as saying ϕ3,-1=1+ϕ1,1.

As our focus has been on integers, it seems reasonable to extend this to consider transformations of the form aϕn,α+b with a,b. That is to say, for a,b, when is aϕn,α+b another ϕm,β?

We can test this by simple substitution:

(aϕn,α+b)2=a2ϕn,α2+2abϕn,α+b2=a2(nϕn,α+α)+2abϕn,α+b2=a2α+(2b+na)(aϕn,α+b)-2b2-nab+b2=(2b+na)(aϕn,α+b)+αa2-b2-nab

Thus if αa2-b2-nab=±1=β then aϕn,α+b=ϕ2b+na,β. Hence we look for solutions of:

αa2-nab-b2=β

with a,b,n and α,β{1,-1}. This is a Diophantine equation. We shall specify a solution by giving (a,b,n,α) and shall always use β to refer to the resulting right hand side (thus β=αa2-nab-b2).

There are some particular cases that will need case-by-case analysis. To see where they lie, let's start with the general case. This is if a,b>2. The point about this restriction is that it allows us to work modulo a and modulo b and still be able to distinguish α and β. In particular, if we work modulo a we obtain -b2βmoda whilst modulo b shows us that αa2βmodb.

Thus a solution requires (a,b) with b2=±1moda, and a2=±1modb. On the other hand, given such a pair, we set β to be such that -b2=βmoda and α such that αa2=βmodb. Then αa2-b2-β0 modulo both a and b. Since b2=±1moda, there is some integer k such that b2+ka=1 whence gcd(a,b)=1. Thus as αa2-b2-β0 modulo both a and b, it must be that ab divides αa2-b2-β. Hence there is some n such that αa2-nab-b2=β.

This argument still basically holds if either of a or b is 2 except that when constructing the solution from the pair (a,b) then we get more than one solution because, say, -b2βmod2 is not enough to specify β from b. As we will see, though, the number of cases with one of a or b being 2 is sufficiently small that it is easier to consider these as simply special cases.

We shall return to this case after analysing the special cases.

The special cases are when we don't have a,b>2. The first observation is that if (a,b,n,α) is a solution then changing the sign of any two of (a,b,n) also yields a solution. So in the search for solutions, we can assume that a,b0. Another useful symmetry of solutions is that if (a,b,n,α) is a solution then so is (b,a,-αn,α). Thus it is sufficient to consider the cases a=0, a=1, and a=2, and for each we can assume that ba.

If a=0 then we have -b2=β, whence β=-1 and b=±1. Substituting in, we get ϕ2,-1=1 and ϕ-2,-1=-1. Swapping a and b, we get ϕn,α=ϕn,α and -ϕn,α=ϕ-n,α.

Let's now deal with the case of a=1. Then α-nb-b2=β. If also b=1 then rearranging gives n=α-1-β. Similarly, if b=2 we get 2n=α-4-β (note that α-β is always even). In these special cases, we find the following identities:

ϕ-1,-1+1=ϕ1,-1ϕ-3,-1+1=ϕ-1,1ϕ1,1+1=ϕ3,-1ϕ-1,1+1=ϕ1,1ϕ-2,-1+2=ϕ2,-1ϕ-3,-1+2=ϕ1,1ϕ-1,1+2=ϕ3,-1ϕ-2,1+2=ϕ2,1

(Note that ϕ1,-1 and ϕ-1,-1 are not real.) Our original observation that ϕ3,-1=ϕ1,1+1 is in this list.

Swapping a and b yields some new identities (and repeats some old ones, not shown):

2ϕ-2,-1+1=ϕ-2,-12ϕ-3,-1+1=ϕ-4,12ϕ1,1+1=ϕ4,12ϕ2,1+1=ϕ6,-1

For b>2 we can get some insight from working modulo b. In particular (still with a=1), working modulo b shows that α=β leaving us with b2+nb=0. As b0, we must have b=-n and this yields:

ϕn,α-n=ϕ-n,α

Combined with the relationship between ϕ-n,α and ϕn,α, this is a restatement of the fact that the two values for ϕn,α add up to n. Swapping a and b yields the intriguing identity:

-nϕn,α+1=ϕαn2+2,-1

The last special case is if a=2. Then 4α-2nb-b2=β. We cannot have b=2 here, so we assume that b>2. Working modulo b, we have 4αβmodb. Since α,β=±1, this means that ±4±1modb and thus b=3 or b=5.

If b=3 then α=β and 3-α=-2n, so either α=1 and n=-1 or α=-1 and n=-2.

If b=5 then α=-β and 5-α=-2n, so either α=1 and n=-2 or α=-1 and n=-3.

Thus we have (with swaps):

2ϕ-1,1+3=ϕ4,12ϕ-2,-1+3=ϕ2,-12ϕ-2,1+5=ϕ6,-12ϕ-3,-1+5=ϕ4,13ϕ1,1+2=ϕ7,-13ϕ-2,-1+2=ϕ-2,-15ϕ2,1+2=ϕ14,15ϕ-3,-1+2=ϕ-11,1

4 General Relations

Let us return to the general case where a,b>0. Here we look for a,b with a2±1modb and b2±1moda. Also, using the symmetry it is enough to consider ba.

Not being a number theorist, the obvious step is to turn to a computer. It is simple to write a function in Sage that does a search up to a specified limit for such pairs and prints the resulting solution.

The n0 and n-2 skip some trivial results. In the range 2ba100, the results are below.

3ϕ-1,1+5=ϕ7,-115ϕ-4,-1+56=ϕ52,-13ϕ-3,-1+8=ϕ7,-117ϕ4,1+4=ϕ76,13ϕ-3,1+10=ϕ11,-117ϕ-4,1+72=ϕ76,14ϕ-4,-1+15=ϕ14,-121ϕ-3,-1+8=ϕ-47,-14ϕ-4,1+17=ϕ18,-121ϕ1,1+13=ϕ47,-15ϕ2,1+2=ϕ14,121ϕ-1,1+34=ϕ47,-15ϕ1,1+3=ϕ11,121ϕ-3,-1+55=ϕ47,-15ϕ-1,1+8=ϕ11,124ϕ-5,-1+5=ϕ-110,-15ϕ-3,-1+13=ϕ11,126ϕ5,1+5=ϕ140,15ϕ-5,-1+24=ϕ23,-129ϕ2,1+12=ϕ82,15ϕ-5,1+26=ϕ27,-133ϕ3,1+10=ϕ119,-16ϕ-6,-1+35=ϕ34,-134ϕ-3,-1+13=ϕ-76,16ϕ-6,1+37=ϕ38,-134ϕ1,1+21=ϕ76,17ϕ-7,-1+48=ϕ47,-134ϕ-1,1+55=ϕ76,17ϕ-7,1+50=ϕ51,-134ϕ-3,-1+89=ϕ76,18ϕ-3,-1+3=ϕ-18,-135ϕ-6,-1+6=ϕ-198,-18ϕ1,1+5=ϕ18,-137ϕ6,1+6=ϕ234,18ϕ-1,1+13=ϕ18,-148ϕ-7,-1+7=ϕ-322,-18ϕ-3,-1+21=ϕ18,-150ϕ7,1+7=ϕ364,18ϕ-8,-1+63=ϕ62,-155ϕ-3,-1+21=ϕ-123,-18ϕ-8,1+65=ϕ66,-155ϕ1,1+34=ϕ123,-19ϕ-9,-1+80=ϕ79,-155ϕ-1,1+89=ϕ123,-19ϕ-9,1+82=ϕ83,-156ϕ-4,-1+15=ϕ-194,-110ϕ3,1+3=ϕ36,163ϕ-8,-1+8=ϕ-488,-110ϕ-3,1+33=ϕ36,165ϕ8,1+8=ϕ536,110ϕ-10,-1+99=ϕ98,-170ϕ2,1+29=ϕ198,-112ϕ2,1+5=ϕ34,-172ϕ4,1+17=ϕ322,-113ϕ-3,-1+5=ϕ-29,180ϕ-9,-1+9=ϕ-702,-113ϕ1,1+8=ϕ29,182ϕ9,1+9=ϕ756,113ϕ-1,1+21=ϕ29,189ϕ-3,-1+34=ϕ-199,113ϕ-3,-1+34=ϕ29,189ϕ1,1+55=ϕ199,115ϕ-4,-1+4=ϕ-52,-199ϕ-10,-1+10=ϕ-970,-1

5 The Golden Ratio

What is particularly interesting are those involving ϕ1,1 since this is the original golden ratio. Running up to 1000, we have:

5ϕ1,1+3=ϕ11,18ϕ1,1+5=ϕ18,-113ϕ1,1+8=ϕ29,121ϕ1,1+13=ϕ47,-134ϕ1,1+21=ϕ76,155ϕ1,1+34=ϕ123,-189ϕ1,1+55=ϕ199,1144ϕ1,1+89=ϕ322,-1233ϕ1,1+144=ϕ521,1377ϕ1,1+233=ϕ843,-1610ϕ1,1+377=ϕ1364,1987ϕ1,1+610=ϕ2207,-1

Unsurprisingly, the numbers appearing are the Fibonacci and Lucas numbers.

Taking two of these at random and substituting them in to the formulae for cos and sin, we see that:

2sinilog(987ϕ+610)=2207i,2cosilog(610ϕ+377)=1364.

More generally, the obvious conjecture from this is that

Fnϕ+Fn-1=ϕLn,(-1)n

(starting with F0=0 and F1=1).

Let us look for more patterns. If we take those involving ϕ2,1 we get the following list (again, looking for a,b1000):

5ϕ2,1+2=ϕ14,112ϕ2,1+5=ϕ34,-129ϕ2,1+12=ϕ82,170ϕ2,1+29=ϕ198,-1169ϕ2,1+70=ϕ478,1408ϕ2,1+169=ϕ1154,-1985ϕ2,1+408=ϕ2786,1

Again, the coefficients on the left follow a kind of Fibonacci pattern, only this time the recurrence seems to be cn=2cn-1+cn-2. The numbers in the sequence on the right are the Companion Pell numbers (OEIS A002203).

This investigation suggests that if we have found a solution (a,b,n,1), then we get a new solution with (an+b,a,n,1). A little more investigation suggests that this generalises to arbitrary α so that from (a,b,n,α) we should get another solution with (α(an+b),a,n,α,). We can prove this:

α(α(an+b))2-nα(an+b)a-a2=α(a2n2+2anb+b2)-αa2n2-αnab-a2=αanb+αb2-a2=-α(αa2-nab-b2)=-αβ{1,-1}.

This provides a way to generate a lot of relations because we have obvious starting points: (1,0,n,α). The first few terms are:

(1,0,n,α)ϕn,α=ϕn,α(αn,1,n,α)αnϕn,α+1=ϕ2+n2,-1(n2+α,αn,nα)(n2+α)ϕn,α+αn=ϕ3αn+n3,α

The second of these was the "intriguing identity".

The sequence of coefficients is a generalised Fibonacci sequence. The initial data is the pair (n,α) and the resulting sequence is formed by the recurrence:

ak(n,α)=α(nak-1(n,α)+ak-2(n,α)),a1(n,α)=1,a0(n,α)=0.

The kth relation has right hand side ϕm,β where β=-(-α)k and m=2ak-1(n,α)+nak(n,α). This simplifies as:

m=2ak-1(n,α)+nak(n,α)=ak-1(n,α)+(ak-1(n,α)+nak(n,α))=ak-1(n,α)+αak+1(n,α)

Thus our resulting identity is:

ak(n,α)ϕn,α+ak-1(n,α)=ϕak-1(n,α)+αak+1(n,α),-(-α)k.

Note that we can run the recurrence backwards. The one before (a,b,n,α) is (b,αa-bn,n,α). In particular, the one before (1,0,n,α) is (0,α,n,α) and the one before that is (α,-αn,n,α) which becomes

αϕn,α-αn=ϕ-αn,α.

Here's the list for (n,α)=(-3,-1).

3ϕ-3,-1+1=ϕ-7,-18ϕ-3,-1+3=ϕ-18,-121ϕ-3,-1+8=ϕ-47,-155ϕ-3,-1+21=ϕ-123,-1144ϕ-3,-1+55=ϕ-322,-1377ϕ-3,-1+144=ϕ-843,-1987ϕ-3,-1+377=ϕ-2207,-1

For (n,α)=(-4,-1), it starts as follows:

4ϕ-4,-1+1=ϕ-14,-115ϕ-4,-1+4=ϕ-52,-156ϕ-4,-1+15=ϕ-194,-1209ϕ-4,-1+56=ϕ-724,-1780ϕ-4,-1+209=ϕ-2702,-1

Our special cases give us other starting points. For example, from ϕ-1,-1+1=ϕ1,-1 we actually get a cycle of length 6:

ϕ-1,-1+1=ϕ1,-1+1=ϕ2,-1-ϕ-1,-1=ϕ1,-1-ϕ-1,-1-1=ϕ-1,-1-1=ϕ-2,-1ϕ-1,-1=ϕ-1,-1

Feeding in our other starting points, we find the following. (We ignore any sequences with ϕ±2,-1 since they are ±1).

Thus all of our "special cases" are not special save for one:

ϕ-3,-1+1=ϕ-1,1

However, even this one is not so special since ϕn,α=-ϕ-n,α so we can rewrite it as:

ϕ3,-1-1=ϕ1,1

whereupon it is revealed as a rearrangement of:

ϕ1,1+1=ϕ3,-1.

The point being that if ϕn,α+b=ϕm,β then also ϕ-m,β+b=ϕ-n,α and this gives us another starting point. However, of the various identities with leading term 1 that we found, only ϕ-3,-1+1=ϕ-1,1 actually gives us a new sequence. So it is pretty special, and the original golden ratio, ϕ, retains its spot as the most glittery of the golden numbers.

6 Golden Primes

At this stage, an interesting question is: how many golden numbers do we need to generate the rest? One approach to answering this is to use a sieve method.

There are two sieve methods that could be used. One is to start with the tautologies ϕn,α=ϕn,α and run the generalised Fibonacci generations forwards and backwards and mark all the ϕm,β that are so produced. (To make this efficient, we would first want to prove that if ϕn,α produces ϕm,β then all of the ones that ϕm,β generates are also in the sequence for ϕn,α.)

The problem with this sieve is that although it provides a simple rule for generating relations between the ϕn,α, it doesn't guarantee that we have found all of them.

The other sieve method would be to use our original method for finding relations: look for pairs (a,b) with a2±1modb and b2±1moda. This is guaranteed to produce all the relations, but not necessarily in a useful order for crossing the ϕn,α off our list.

Thus to produce a definite list of the "golden primes", we would need to prove either that all relations could be generated by the Fibonacci method, or that for a given ϕn,α then there was a bound on (a,b) beyond which we couldn't find a relation that produced ϕn,α. Both of these seem reasonable things to investigate further, but this article is already long enough. So we will, instead, run the (a,b)-sieve and cross off all generated ϕn,α that we find.

Note that with the sieve then it is sufficient to continue with our assumption that a,b0. A quick implementation in Sage reveals that the overwhelming majority of pairs (n,α) cannot be generated by others.

7 Golden Conclusion

This was nothing more than a merry jaunt with numbers. It started with a simple observation and led on from there. As promised, nothing deep was discovered but I did stumble upon a nice link from the Golden Numbers to a certain type of Diophantine Equations, a topic that I've heard mentioned from time to time but never found a reason to look more closely at. It might be interesting to use them as a way to learn a bit more about Diophantine Equations.

So while this might not qualify as genuine mathematical gold, it certainly has its sparkle. And to the right mind, iron pyrites is far more interesting than dull old gold.