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        riddles >> putnam exam (pure math) >> (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
        
(Message started by: william wu on Nov 8th, 2006, 2:09pm)
Title: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by william wu on Nov 8th, 2006, 2:09pm
Prove that (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3) is a rational number.

Hint 1: [hide]I got this problem out of a mathematical olympiad book that I've been reading. In this chapter, we want to use the following trick: a^3 + b^3 + c^3 = 3abc if a+b+c = 0. Maybe there are other ways to solve this problem though.[/hide]

Hint 2: [hide]Let x = (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3), and then you have an equation with three terms which add to zero. Apply the identity from the previous hint. Now we are getting close.[/hide]
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by towr on Nov 8th, 2006, 3:09pm
Are you sure you didn't make any typos? Because I get quite irrational results..
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by william wu on Nov 8th, 2006, 6:09pm
Hmm, well there are real cube roots and complex cube roots. If you take the purely real cube roots of each term, and then add them, the result is rational. Maybe this should have been stipulated in the problem. Admittedly I've never quite known under what conditions it's understood that the strictly real roots are being considered.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Icarus on Nov 8th, 2006, 6:25pm
Without going into details, if you were to ask me if that was rational or irrational, I would have guessed rational, for much the same reason that (2+i)1/3 + (2-i)1/3 is real.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Eigenray on Nov 8th, 2006, 7:54pm
If x = a+b, then
x3 = (3ab)x + (a3+b3).
(Conversely, if
x3 - px - q = 0
then we may find a,b, such that
3ab = p, a3+b3=q.
This is [link=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]Cardano's method[/link] for solving the cubic equation.)

Letting w = e2 pi i/3 denote a cube root of unity, notice that the numbers
a+b,  wa + w2b,  and w2a + wb
all satisfy the same cubic equation,
x3 = 3abx + (a3+b3),
and in general the solutions to the equation
x3 = 3wiabx + (a3+b3)
are given by
x = wja + wi-jb,  j=0,1,2.

There are 9 possible values of
(2+sqrt(5))1/3 + (2-sqrt(5))1/3,
depending on which cube roots you take, and they each satisfy one of 3 cubic equations.  But only one of these values, namely [hide]1[/hide], is real, let alone rational.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Barukh on Nov 8th, 2006, 10:11pm

on 11/08/06 at 19:54:23, Eigenray wrote:
If x = a+b, then
x3 = (3ab)x + (a3+b3).

Yes! This immediately gives a method of constructing the radical equal to any prescribed integer.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Michael_Dagg on Nov 9th, 2006, 12:16pm
HA! Fun.

Consider  ((1 + sqrt(5))/2)^3 = 2 + sqrt(5).

Can you take it from there?
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Ken_Wiley on Nov 9th, 2006, 10:11pm
Apparently 2-sqrt(5)=((1-sqrt(5))/2)^3 also.  I guess your point is to replace those two original terms with the new forms, after taking cube roots, and end up with 1 which, is rational of course.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Michael_Dagg on Nov 10th, 2006, 3:26pm

on 11/09/06 at 22:11:38, Ken_Wiley wrote:
I guess your point is to replace those two original terms with the new forms, after taking cube roots, and end up with 1 which, is rational of course.


Well yes. If you are only interesting in the rationals (other any other reals) then
that's all you need and hence, QED.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Ken_Wiley on Nov 10th, 2006, 5:58pm
I have a book about identities and formulas but I don't find these in it. There are many other formulats with the sqr root of 5 in them.  I put that formula in to my TI calculator and it simplified.  Really that confuses me because when I put in the original equation it gave me a complex number.

How did you know that formula?
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Michael_Dagg on Nov 10th, 2006, 6:32pm
It is just one of those things you remember and I 'll bet
you'll remember it now!

By the way, what complex value did your calculator display?
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Barukh on Nov 10th, 2006, 7:15pm

on 11/10/06 at 17:58:16, Ken_Wiley wrote:
How did you know that formula?

The formula can be derived as follows: Write

[a+b sqrt(5)]3 = 2 + sqrt(5)

Expanding and equating the rational and irrational parts, we get:

a3 + 15ab2 = 2,
3a2b + 5b3 = 1.

Multiply the second equation by 2, subtract the first and divide through b3; denoting a/b = z, we get:

z3 - 6z2 + 15z -10 = 0

Note that the sum of the coefficients of this cubic equals 0, implying z = 1.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Ken_Wiley on Nov 10th, 2006, 7:39pm
I am more than slightly lost with what you did there.

Seems that what you did worked up backwards. With the z equation you get the two equations and then the one. But I think I see what you might mean by a formula.

The way I did it was to expand it out and it simplifed very easy.

I guess the best question to ask is if these formulas are known then why would a problem like this appear?  Is it just to test students about their knowledge or what?  Someone slap me but when I look back at the original problem and read its details I must assume that it is complex but along comes a formula that makes it simple, but in between there are complex discussions and at one point the number 1 is mentioned but then the statment says "let alone rational" which means to mean No.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by towr on Nov 13th, 2006, 1:14am
If you have a puzzle with sqrt(5) in it, suspect fibonacci (or at least phi=(1+sqrt(5))/2) is involved. (Unfortunately when I saw this puzzle I didn't set through on that thought)

Of course it's more general than that; you can find the rational value for e.g.
(99 + 70 sqrt(2))^(1/3) + ( 99 - 70 sqrt(2))^(1/3)
in a similar way.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Ken_Wiley on Nov 13th, 2006, 9:15am
Thanks for the tip. I am learning to see the humor an apparent puzzles within puzzles in between the lines on this forum. I wondered if there was a sqrt(2) analogy to sqrt(5) and you mentioned it!

When I first saw ((1+sqrt(5))/2 I thought it looked like something you'd get from the quadratic formula and it is for the equation  x^2 - x - 1.
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by towr on Nov 13th, 2006, 9:58am

on 11/13/06 at 09:15:13, Ken_Wiley wrote:
I wondered if there was a sqrt(2) analogy to sqrt(5) and you mentioned it!
I think it works for every squareroot. But after 5 the most popular choice is 2 :P


Quote:
When I first saw ((1+sqrt(5))/2 I thought it looked like something you'd get from the quadratic formula and it is for the equation  x^2 - x - 1.
Yup, the connection with fibonacci comes from this as well.
The recurrence for the fibonacci sequence is f(0)=0, f(1)=1, f(n) = f(n-1)+f(n-2)
And you find the closed form by solving
rn= rn-1+ rn-2.
Dividing by rn-2 you get the quadratic
r2= r1 + 1, or r2 - r1 - 1 = 0 ,
Taking the two solutions
phi = ((1+sqrt(5))/2 and phi' = ((1-sqrt(5))/2
you have f(n) = (phin - phi'n )/sqrt(5)

(It works in a similar way for other second order recurrences)
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by Eigenray on Nov 13th, 2006, 5:35pm

on 11/10/06 at 19:39:33, Ken_Wiley wrote:
I am more than slightly lost with what you did there.

Seems that what you did worked up backwards. With the z equation you get the two equations and then the one. But I think I see what you might mean by a formula.

The way I did it was to expand it out and it simplifed very easy.

It's easy to check that ([1+sqrt(5)]/2)3 = 2 + sqrt(5), but the problem is, if we don't already know this, how can we compute [2 + sqrt(5)]1/3?  We suppose that the result is of the form a + b sqrt(5), i.e.,

(*)  [a+b sqrt(5)]3 = 2 + sqrt(5)

and we try to find a and b, if they exist (and they might not: for example, (3+sqrt(5))1/3 is not in the form a + b sqrt(5) for any rational a,b).  Assuming a,b are rational, we can expand out (*) and compare the rational and "irrational" parts, giving two equations in the two variables a,b.  We can then deduce the homogeneous equation

a3 + 15ab2 = 2 [3a2b + 5b3],

which, dividing through by b3, and letting z=a/b, becomes

(**) z3 - 6z2 + 15z - 10 = 0.

By inspection, this has the rational root z=1, and since

z3 - 6z2 + 15z - 10 = (z-1)(z2 - 5z + 10),

we can see this is the only rational root.  That is, if (*) is satisfied, then (**) must be also, and since z=a/b is rational, this means z=1, i.e., a=b.  Thus we need to solve

a3 + 15aa2 = 2,
3a2b + 5a3 = 1,

which is satisfied when a=b=1/2.  (The fact that a + b sqrt(5) happens to be the golden ratio is just a coincidence; for example
[38 + 17 sqrt(5)]1/3 + [38 - 17 sqrt(5)]1/3 = [2 + sqrt(5)] + [2 - sqrt(5)] = 4.)


Quote:
I guess the best question to ask is if these formulas are known then why would a problem like this appear?

Like any problem, the solution is only "known" once you've either worked out, or been shown, how to do it.  So one way to solve this problem is to (figure out how to) explicitly compute [2+sqrt(5)]1/3, and then show that the sum in question is 1.


To clarify my previous post:

Let x = u + v, where u3 = r + s sqrt(d),  v3 = r - s sqrt(d).

(So that for the present problem, r=2, s=1, d=5).  Then from the identity
(u+v)3 = u3 + v3 + 3uv(u+v)
we obtain
x3 = 2r + 3(r2-s2d)1/3x.

Assume that r,s,d are rational, and d>0 is not a square.  Then by taking real cube roots u,v, we have that x=u+v and the coefficient uv above are both real.  In general the polynomial

f(x) = x3 + px + q = x3 - 3(r2-s2d)1/3x - 2r

has only one real root, since its discriminant is

D = -4p3 - 27q2 = -4*[-33(r2 - s2d)] - 27[-2r]2 = -108d s2 < 0.

(Unless s=0, in which case the roots will be 2r1/3, -r1/3, -r1/3, the first of which corresponds to taking u=v real.)

So given r,s, the value of x (obtained by taking real roots u,v) is the (unique) real root of the above polynomial f.  It is rational iff f has a rational root, and since f is monic, the only possible rational roots are integer divisors of q = -2r.  In the case r=2, s=1, d=5, we get f(x) = x3 + 3x - 4 = 0, which does have the rational root x=1.


Now, to compare the two methods.  Clearly, if we can find rational a,b such that [r + s sqrt(d)]1/3 = a + b sqrt(d), then x = 2a will also be rational (this is where we use d is not a square).

On the other hand, suppose x = u+v = 2a is rational, with u,v as above, with r,s rational.  Now, any two numbers with sum 2a can be written in the form a +/- b sqrt(d) for some number b.  So for some b,
u = [r + s sqrt(d)]1/3 = a + b sqrt(d)
v = [r - s sqrt(d)]1/3 = a - b sqrt(d).
Cubing both sides, and taking sums and differences gives
2r = 2[a3 + 3dab2]
2s sqrt(d) = 2[3a2b + db3] sqrt(d).
Since a,r,d are rational, the first equation shows b2 is rational (unless x=r=a=0, which is boring).  Using this fact in the second equation then shows b itself is rational.  That is, if x as above is rational, then we can find a cube root of r + s sqrt(d) of the form a + b sqrt(d), with a,b, rational.  (This isn't particularly obvious.  For example, this says that if u,v are distinct real numbers with u3, v3, and u+v all rational, then u and v must both be rational.)

So x is rational iff r=0 or [r + s sqrt(d)] is a cube in Q(sqrt(d)).  And x is rational iff the cubic
f(x) = x3 - 3(r2-ds2)1/3x - 2r
has a rational root.


Problems like this are related to Cardano's method.  For example, suppose we want to solve the equation
x3 + px + q = x3 + 3x - 4 = 0.
Using Cardano's method, we get
x = u + v = (-q/2 + sqrt(-D/108))1/3 + (-q/2 - sqrt(-D/108))1/3
= (2+sqrt(5))1/3 + (2-sqrt(5))1/3,
where the cube roots are chosen so that uv = -p/3 = -1.  But it's not at all obvious from this "closed form" expression that one of the roots is rational, whereas we can see that x=1 is a solution to the original equation.

The situation is even worse when the polynomial has 3 real roots.  For example, the solutions to
x3 - 7x + 6 = 0
are "given" by
x = [-3 + 10i/(3sqrt(3))]1/3 + [-3 - 10i/(3sqrt(3))]1/3,
where the cube roots are chosen to have product 7/3.  Now it's not clear that any roots are rational, but we can easily see from the original equation that they are 1,2, and -3.

[If we have an irreducible cubic equation with 3 real roots, however, Cardano's method is in a sense the best we can do, because it's a theorem that then none of the roots can be expressed in terms of real radicals.  That is, the roots are not contained in any root extension consisting of subfields of R.]
Title: Re: (2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3)
Post by towr on Nov 14th, 2006, 12:05am

on 11/13/06 at 17:35:42, Eigenray wrote:
(The fact that a + b sqrt(5) happens to be the golden ratio is just a coincidence
Except that we're dealing with a puzzle, not a random problem ;)


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