View Full Version : Math Poser

Luckmeister

2010-Oct-02, 06:27 PM

Last night a fellow presented me with a math poser. He said that a^2 + b^2 = c^2 is a common and fairly easy to solve equation but he challenged me to solve a^3 + b^3 = c^3. He said there is only one answer but he didn't know what it was.

I asked if a and b are integers or if fractions would have to be used and he said he didn't know. I considered writing a program to find the solution but don't have the time right now.

Is anyone familiar with this?

Mike

pzkpfw

2010-Oct-02, 08:09 PM

a3 + b3 = c3

Assuming they have to all be integers: a, b and c are all 1?

Or, a = 0 and c = b.

Luckmeister

2010-Oct-02, 09:51 PM

Ah yes, I hadn't thought of that. Thanks! :lol:

HenrikOlsen

2010-Oct-02, 09:59 PM

Definitely integers, also must be non-negative positive integers, otherwise such trivial solutions as 0^3+0^3=0^3 and -1^3+1^3=0^3 would be allowed.

It's easy to prove that there are infinitely many integer solutions to a^2 + b^2 = c^2, simple proof is that if you take any odd a>1 then b=(a^2-1)/2 and c=(a^2+1)/2 will give a solution, easily seen by:

a=2n+1 => a^2=4n^2+4n+1

b=(a^2-1)/2=(4n^2+4n+1-1)/2=2n^2+2n => b^2=4n^4+8n^3+4n^2

c=(a^2+1)/2=(4n^2+4n+1+1)/2=2n^2+2n+1 => c^2=4n^4+8n^3+8n^2+4n+1

(4n^2+4n+1)+(4n^4+8n^3+4n^2)=(4n^4+8n^3+8n^2+4n+1)

This won't give all solutions but does give infinitely many. This was BTW a proof I made myself some 20 odd years ago when I was playing around with the formula.

I remember finding a similar constructor for a different infinite set of triples, starting with b=2^2n, but I can't remember the rest of the details now.

The one with ^3's is part of Fermat's Last Theorem (http://en.wikipedia.org/wiki/Fermat's_Last_Theorem), though it technically didn't become one until proven late last century.

Your friend was teasing you, it has been proven that it is impossible to solve a^n + b^n = c^n for non-zero positive integers a, b and c when n>2, though it took quite a lot of effort to do so.

Jens

2010-Oct-03, 04:43 AM

a3 + b3 = c3

Assuming they have to all be integers: a, b and c are all 1?

If a, b, and c are all 1, then don't you get:

1 + 1 = 1 ?

Hmmm.

pzkpfw

2010-Oct-03, 05:06 AM

If a, b, and c are all 1, then don't you get:

1 + 1 = 1 ?

Hmmm.

Er, that depends on what value you use for 1.

Ivan Viehoff

2010-Oct-04, 04:05 PM

Proving that there is no solution in integers for a^3 + b^3 = c^3 with a, b, c > 0, is within the range of high school maths, and is quite a short easy proof. Here: http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html This proof is due to Euler. Euler also had a previous mistaken proof. The quartic case is a whole lot trickier, but I think still within the range of undergrad number theory. Fermat himself proved the n=3 and n=4 case. Beyond that, things get really nasty. But before Wiles completed the entirely general proof for all exponents (during which he stood on the shoulders of some giants - Taniyama deserves a special mention), I think the specific cases for all exponents up to several million had been demonstrated.

Incidentally, in the popular book The Girl who Played with Fire by Stieg Larssen (which I thought was substantially trashier than TGWTDT, but not for the following reason), the author seems to think that the cubic case is Fermat's Last Theorem. He also seems to think that Wiles used a computer in its proof. I guess he confused it with the Four Colour theorem. He also made some nonsensical comment about a chess game. A former schoolmate of mine had a prize-winning novel published with a similar chess-related solecism. Why does this happen? I suppose if the IWC report on global warning had sufficiently poor QA that the obviously ridiculous 2035 Himalayan glacier melt claim remained, we shouldn't be surprised.

grapes

2010-Oct-04, 04:16 PM

Last night a fellow presented me with a math poser. He said that a^2 + b^2 = c^2 is a common and fairly easy to solve equation but he challenged me to solve a^3 + b^3 = c^3. He said there is only one answer but he didn't know what it was.One night, after thinking about this, I tried to find a positive integers solution to a^3 + b^3 + c^3 = d^3, while lying in bed. I was able to find one before I went to sleep.

Why does this happen?Human nature. :)

I suppose if the IWC report on global warning had sufficiently poor QA that the obviously ridiculous 2035 Himalayan glacier melt claim remained, we shouldn't be surprised.Stop that, right now! Do not sully a pristine math thread with global warming discussion.

pghnative

2010-Oct-12, 01:21 PM

One night, after thinking about this, I tried to find a positive integers solution to a^3 + b^3 + c^3 = d^3, while lying in bed. I was able to find one before I went to sleep. Only one? I guess you fell asleep before you made it up to eights.

Agemegos

2010-Oct-16, 01:53 AM

Last night a fellow presented me with a math poser. He said that a^2 + b^2 = c^2 is a common and fairly easy to solve equation but he challenged me to solve a^3 + b^3 = c^3. He said there is only one answer but he didn't know what it was.

I asked if a and b are integers or if fractions would have to be used and he said he didn't know. I considered writing a program to find the solution but don't have the time right now.

Is anyone familiar with this?

Yep, it's the n=3 case of Fermat's Last Theorem.

It happens that there are no solutions for a, b, and c all integers, except for trivial ones in which either a or b is zero. It follows from this that there are no solutions for a, b, and c all rationals, because if there were you could multiply through by the cube of any common multiple of the denominators of a, b, and c and get a solution in integers.

It should be perfectly obvious that there are infinitely many solutions in the reals, because the reals are closed under multiplication and addition, and every real has a cube root that is also real. That is, you can find d = cuberoot (a3 + b3) for any reals a and b.

It also ought to be obvious that although there might be no solutions and there might be infinite solutions, there can't be only one solution. To see this, assume that there is a solution

a3 + b3 = c3

Multiply both sides by d3, for any d.

d3.(a3 + b3) = d3.c3

Apply the distributive law of multiplication over division.

d3.a3 + d3.b3 = d3.c3

Apply the associative law of multiplication

(d.a)3 + (d.b)3 = (d.c)3

Conclusion: if (a, b, c) is a solution, (d.a, d.b, d.c) is also a solution for any d in the field.

Agemegos

2010-Oct-16, 02:00 AM

It also ought to be obvious that although there might be no solutions and there might be infinite solutions, there can't be only one solution.

Umm. Except in a finite field. Integer arithmetic modulo three might have only one distinct non-trivial solution.

grapes

2010-Oct-16, 03:45 AM

Umm. Except in a finite field. Integer arithmetic modulo three might have only one distinct non-trivial solution.Definitely no chance of having an infinite number of solutions, in a finite field. :)

HenrikOlsen

2010-Oct-16, 09:11 AM

Definitely no chance of having an infinite number of solutions, in a finite field. :)

The equivalent to infinity in a finite field would be to have solutions for all numbers in the field.

peter eldergill

2010-Oct-16, 01:29 PM

I'm not sure how common the knowledge is that Fermat's last theorem was previously proved for specific cases (I"m not sure offhand beyond n=3 and 4, but I think there are more cases)

I certainly didn't know until fairly recently (last 10 years or so) and I think I may have read about it in an article somewhere

Pete

Agemegos

2010-Oct-16, 10:53 PM

The equivalent to infinity in a finite field would be to have solutions for all numbers in the field.

I guess so. I was just thinking that d = 0 is trivial and d = 1 does not produce a distinct solution, so my proof that there cannot be only one distinct non-trivial solution does not hold when 0 and 1 are the only possible values of d.

It's funny applying Fermat's Last Theorem to fields other than the natural numbers under multiplication over division. In Boolean Algebra every pair (a, b) has a solution c for every n, giving four solutions for each n: but three of those are trivial. 1n ⊕ 1n = 1n is a non-trivial solution for all n. In Modulus Two arithmetic, 1n + 1n = 0n {mod 2} is a non-trivial solution for all n. But then, exponentiation is trivial in those fields. xn = x, for all x.

grapes

2010-Oct-17, 08:02 AM

I dunno, I would even count (A)n + (-A)n = (0)n as a trivial solution.

Agemegos

2010-Oct-19, 01:58 AM

Harsh!

Ken G

2010-Oct-19, 02:49 AM

I dunno, I would even count (A)n + (-A)n = (0)n as a trivial solution.Can't be too trivial-- it only works for odd n!

grapes

2010-Oct-19, 10:55 AM

When it comes to Fermat's Last Theorem, odd n are the only ones that matter. At least, in my lifetime (which has been a good while). :)

Agemegos

2010-Oct-20, 04:21 AM

Can't be too trivial-- it only works for odd n!

Depends on the field. In modulus 2 arithmetic it holds for all integer n, because 1 + 1 = 0 {mod 2}.

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