Does the Equation a3 + b3 = c3 Have Any Geographical Significance?

[Loren Booda]

Does the equation

a³ + b³ = c³

(where a, b and c are constants) have any general geometrical significance?

 

[John Creighto]

Does the equation

a³ + b³ = c³

(where a, b and c are constants) have any general geometrical significance?

Not sure. What about if we give a triangle depth and do something like the Pythagoras theorem but use area instead of length?

 

[tiny-tim]

What about if we give a triangle depth and do something like the Pythagoras theorem but use area instead of length?

No way!

I don't think cubes ever arise in geometry.

Maybe in calculus, and in some branches of physics, but not in geometry.

 

[jostpuur]

Are you interested about values a,b,c\in\mathbb{N} or \in\mathbb{R}?

With natural numbers, there is an equivalent problem with small cubes of fixed size. Suppose you have c^3 small cubes, piled into a one bigger cube. Can you take these c^3 small cubes, and form two new big cubes that contain precisely all these small cubes? The Fermat's last theorem says that you cannot.

With real numbers, an equation x^3+y^3=1 describes a surface of a L^p ball with value p=3. Frankly, I still have not understood why real analysis deals so much with L^p spaces. I don't know what's their significance, yet

 

[HallsofIvy]

No way!

I don't think cubes ever arise in geometry.

Maybe in calculus, and in some branches of physics, but not in geometry.

Well, so much for solid geometry!

 

[John Creighto]

No way!

I don't think cubes ever arise in geometry.

Really? How about volume?

 

[tiny-tim]

oh, those cubes! ​

™​

 

[John Creighto]

oh, those cubes! ​

™​

Or spheres.

 

[tiny-tim]

… or Russian dolls! … ​

 

[Bananaman3579]

Ok tell me if there's a problem with this but I think this is the only way to solve it.
cube root 5^3+cube root 5^3= cube root 10^3
the cube root just cancels out the cubing so the a and b variables need to be the same and the c variable needs to be double the a or b.
I don't see what use this would have but that's the only way I've found to solve it.

 

[Bananaman3579]

I'm not bananaman, but a friend and I think his reasoning is flawed somewhere but the equation seems reasonable. I still believe that the equation is universally unsolvable, basically forbidden. WC

 

[tiny-tim]

… daylight come, and I want to go home …

How many Bananamen are there?

…Where is Talleyman when we need him! … ​

 

[Alex48674]

There is no solution. No 2 (natural number) cubes add up to another cube. In fact no solutions exist in x^n + y^n =z^n, where n is any number greater then 2, as stated by Fermat's last theorem (as proved by Andrew Wiles).

bananana you can't cancel out like this. One way to show this is wrong is

5^3 +5^3= 10^3

125+125=1000

250=1000

and this is clearly not true.

 

[Bananaman3579]

I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.

 

[Bananaman3579]

o and the "other" bananaman was just my friend using my account

 

[Alex48674]

I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.

Ok sorry I didn't see that bit, but regardless it's the same.

lets say you can cancel out the cube then

a=x^1\2
b=y^1/2
c=z^1/2

then a^3 +b^3= c^3

according to you a+b=c

so let's say 1+1=2

then plug back into the formula you get 1+1=8 which is not true

the cube rrot is irrelevantly

 

[Bananaman3579]

ok so I actually did this on a calculator, feel free to do so yourself, and the equation works. Now the problem i see with it is that it takes away from the "spirit" of the equation in the sense that it is just canceling out the main part of the equation and therefore basically destroying it. It reminds me of those verbal equations kids would say like, give me your favorite number then multiply by 2 add this and that, and the number would end up being the original through inverse operations. Basically i proved that there is a way to solve it but its pointless.

 

[Bananaman3579]

ok you might also have missed the fact that the number that had a cube root was the variable, a= cube root of one b= cube root of 1 c= cube root of 2. then plug that into the equation.

 

[Bananaman3579]

… or Russian dolls! … ​

or when dividing by zero !O KNOW BLACK HOLE!...fixzzle

 

[Alex48674]

Sorry I had to go and finish, I was going to say:

a=x^1\3
b=y^1/3
c=z^1/3 This takes care of you cube root problem

then a^3 +b^3= c^3

according to you a+b=c

so let's say 1+1=2

then plug back into the formula you get 1+1=8 which is not true

Then if you want to convert back to your cubed root variables you get

1 + 1= 512

My point is yes, for yours it works because basically what you are doing is picking numbers that will work for your method, but you can not apply this method to everything because it will not work for all cases in which a +b=c would also work for the cube of each.

Also the reason the cube root is irrelevantly is because every number has a cube root, so if we have the number a, the cube root could be x ( cube root a=x) and then simplify the problem so that x^3 + y^3=z^3, and I I showed this does not work. Look up fermats last theorem it disproves it all.

 

[Bananaman3579]

Ok I think your still missing something. First things first can we at least agree there is no application for this no matter what, lol. Moving on to the utterly pointless. You switched your variables a bit. Looking at what you said in your most recent post x+y=z not a+b=c because then the variables get messed up with the cube root.

humor me here just a little more.
get a calculator

then plug in ([cube root 2]^3)+([cube root 2]^3)and that should equal 4.
you can also replace the 2s with any two numbers and the answer will be the two numbers added together.

I did look up fermats last theorum and it was hard to follow especially because I couldn't actually find the proof.

He also doesn't state that the power can't be negative which would send the equation spiraling into the abyss.

 

[mathwonk]

have you heard of elliptic curves? there is a lot of beautiful geometry associated with cubic plane curves like this.

 

[Alex48674]

Sorry Banana I miss read you a bit:

But I figured out where it is wrong

So basically what it is is that the problem is a^3+b^3=c^3

but you changed it to cube root a^3 +cube rootb^3 =cube rootc^3
which is equal to a+b=c, so if what you say is right you would be able to use any terms for this second equation and it would fit the third
so let's say 1+1=2

then plug in you get 1+1=8... doesn't work. What you have done is changed the problem completely, you can not justify taking the cubic roots of all the terms and still call it the same problem. So in short it is not a solution. In fact there is no solution, it is impossible to find one.

 

[Alex48674]

have you heard of elliptic curves? there is a lot of beautiful geometry associated with cubic plane curves like this.

i have heard of it, and I want to do it, but sadly I am nowhere near high enough maths for it ='[, only in 9th grd algII

Oh and banana try searching fermats last theorem n=3, you might get some results, this was one of the first solutions disprove in the theorem, and it was thought that by elimination every prime and n=4,6 then it would be solved, but there are an infinite number of primes. In the end it was solved with a proof connecting an elliptic curve conjure with modular forms with fermats theorm.

 

[Mute]

Sorry Banana I miss read you a bit:

But I figured out where it is wrong

So basically what it is is that the problem is a^3+b^3=c^3

but you changed it to cube root a^3 +cube rootb^3 =cube rootc^3
which is equal to a+b=c, so if what you say is right you would be able to use any terms for this second equation and it would fit the third
so let's say 1+1=2

then plug in you get 1+1=8... doesn't work. What you have done is changed the problem completely, you can not justify taking the cubic roots of all the terms and still call it the same problem. So in short it is not a solution. In fact there is no solution, it is impossible to find one.

As far as I can tell Bananaman was simply trying to provide a set of numbers that solved the equation a^3 + b^3 = c^3. He never made any reference to trying to find natural numbers that solved the problem. Hence, for any x, y and z such that x = a^3, y=b^3 and z=c^3, the equation is solved. You've been assuming he's been trying to find natural number solutions and hence have been telling him there is no solution, but if he's just trying to find a triple of real numbers that satisfies the equation, then he's done that and telling him there's no solution is bound to just be confusing.

 

[Loren Booda]

Thank you, mathwonk.

I believe elliptic curves may be near impossible to solve analytically.

The Wiki article on elliptic curves quotes my old professor, Serge Lang.

 

[Bananaman3579]

Yeah I might be in a little over my head here but all I was finding was a set of numbers that would make that equation true, NOT a solution that would work infinitly. basically what I'm saying is that you can take any two numbers, take th cube root of them and plug that into A and B. thus making the answer to the equation as the addition of the first two original numbers.

A^3+B^3=C^3

A= cube root 5
B= cube root 9
C= cube root 14

plug them in and the equation is true. CALCULATOR WILL PROVE ALL...or explode

 

[Alex48674]

As far as I can tell Bananaman was simply trying to provide a set of numbers that solved the equation a^3 + b^3 = c^3. He never made any reference to trying to find natural numbers that solved the problem. Hence, for any x, y and z such that x = a^3, y=b^3 and z=c^3, the equation is solved. You've been assuming he's been trying to find natural number solutions and hence have been telling him there is no solution, but if he's just trying to find a triple of real numbers that satisfies the equation, then he's done that and telling him there's no solution is bound to just be confusing.

Yea I see what you mean, I looked through and I never saw anything about natural numbers, but still the whole point of this problem is the natural number part, other wise it is a straight forward equation. I think whoever brought this up though meant natural numbers, but forgot to mention it.

If we assume natural numbers though, then there are no solutions. By the way banana I can't seem to remember why, but I think there is something logically wrong with canceling out roots and powers, although in practice it does end up working most of the time. Can't seem to remember why at the moment though.

 

[Mute]

Yeah I might be in a little over my head here but all I was finding was a set of numbers that would make that equation true, NOT a solution that would work infinitly. basically what I'm saying is that you can take any two numbers, take th cube root of them and plug that into A and B. thus making the answer to the equation as the addition of the first two original numbers.

A^3+B^3=C^3

A= cube root 5
B= cube root 9
C= cube root 14

plug them in and the equation is true. CALCULATOR WILL PROVE ALL...or explode

Yes, that's true, but the reason the people found the problem interesting was that Fermat supposedly had a theorem (which never got written down) that showed that the equation

a^n + b^n = c^n

has no solutions such that a, b and c are Natural Numbers (integers greater than zero) for n > 2. There are of course an infinite number of possible triples (a,b,c) that do solve the equation: if you set c = 5, for example, and plot y = (5^3-x^3)^(1/3) you'll get a curve, points along which will satisfy x^3 + y^3 = 5^3, but not both of x and y will be natural numbers, by the theorem (which was not actually proven until recently by Wiles).

 

[Alex48674]

Yes, that's true, but the reason the people found the problem interesting was that Fermat supposedly had a theorem (which never got written down) that showed that the equation

a^n + b^n = c^n

has no solutions such that a, b and c are Natural Numbers (integers greater than zero) for n > 2. There are of course an infinite number of possible triples (a,b,c) that do solve the equation: if you set c = 5, for example, and plot y = (5^3-x^3)^(1/3) you'll get a curve, points along which will satisfy x^3 + y^3 = 5^3, but not both of x and y will be natural numbers, by the theorem (which was not actually proven until recently by Wiles).

Exactly =]

 

[Bananaman3579]

wasnt there an error in Wiles work too?

 

[CRGreathouse]

wasnt there an error in Wiles work too?

In the original version, yes. It took a few other mathematicians many months to help Wiles fix the mistakes.

 

[Diffy]

Thank you, mathwonk.

I believe elliptic curves may be near impossible to solve analytically.

The Wiki article on elliptic curves quotes my old professor, Serge Lang.

Serge Lang RIP.

Wow I am very jealous that you had the opportunity to learn from one of the best! Were you his student at Yale? His textbooks are classics and a mandatory part of any algebraist collection. Was he as good of a teacher as he was an author?

 

[Math Jeans]

I would think that if you are going to introduce cubing into the specific equation, then you would need an extra variable as it would effect a higher dimension.

Mainly: a^3+b^3+c^3=d^3

 

[Loren Booda]

Good idea, Math Jeans!

 

[iamafigment]

As I understand it, you are looking for values for a,b, and c where
a+b=c AND a^3 + b^3 = c^3

How about this?
a=cube rt 1
b=cube rt 1
c=cube rt 2

 

[robert Ihnot]

However, it is clearly the case that three cubes can equal a cube. The simplist case is:
3^3+4^3+5^3=6^3

However as a generalization from 3^2+4^2=5^2, it does not seem to stretch very far.

 

[CRGreathouse]

3^3+4^3+5^3 = 6^3
1^3+6^3+8^3 = 9^3
6^3+8^3+10^3 = 12^3
2^3+12^3+16^3 = 18^3
9^3+12^3+15^3 = 18^3
3^3+10^3+18^3 = 19^3
7^3+14^3+17^3 = 20^3
12^3+16^3+20^3 = 24^3
4^3+17^3+22^3 = 25^3
3^3+18^3+24^3 = 27^3
18^3+19^3+21^3 = 28^3
11^3+15^3+27^3 = 29^3
15^3+20^3+25^3 = 30^3
4^3+24^3+32^3 = 36^3
18^3+24^3+30^3 = 36^3
6^3+20^3+36^3 = 38^3
14^3+28^3+34^3 = 40^3
2^3+17^3+40^3 = 41^3
6^3+32^3+33^3 = 41^3
21^3+28^3+35^3 = 42^3
16^3+23^3+41^3 = 44^3
5^3+30^3+40^3 = 45^3
3^3+36^3+37^3 = 46^3
27^3+30^3+37^3 = 46^3
24^3+32^3+40^3 = 48^3
8^3+34^3+44^3 = 50^3
29^3+34^3+44^3 = 53^3
6^3+36^3+48^3 = 54^3
12^3+19^3+53^3 = 54^3
27^3+36^3+45^3 = 54^3
36^3+38^3+42^3 = 56^3
9^3+30^3+54^3 = 57^3
15^3+42^3+49^3 = 58^3
22^3+30^3+54^3 = 58^3
21^3+42^3+51^3 = 60^3
30^3+40^3+50^3 = 60^3
7^3+42^3+56^3 = 63^3
33^3+44^3+55^3 = 66^3
22^3+51^3+54^3 = 67^3
36^3+38^3+61^3 = 69^3
7^3+54^3+57^3 = 70^3
14^3+23^3+70^3 = 71^3
8^3+48^3+64^3 = 72^3
34^3+39^3+65^3 = 72^3
36^3+48^3+60^3 = 72^3
12^3+51^3+66^3 = 75^3
38^3+43^3+66^3 = 75^3
12^3+40^3+72^3 = 76^3
31^3+33^3+72^3 = 76^3
39^3+52^3+65^3 = 78^3
28^3+56^3+68^3 = 80^3
9^3+54^3+72^3 = 81^3
25^3+48^3+74^3 = 81^3
4^3+34^3+80^3 = 82^3
12^3+64^3+66^3 = 82^3
19^3+60^3+69^3 = 82^3
28^3+53^3+75^3 = 84^3
42^3+56^3+70^3 = 84^3
54^3+57^3+63^3 = 84^3
50^3+61^3+64^3 = 85^3
20^3+54^3+79^3 = 87^3
26^3+55^3+78^3 = 87^3
33^3+45^3+81^3 = 87^3
38^3+48^3+79^3 = 87^3
21^3+43^3+84^3 = 88^3
25^3+31^3+86^3 = 88^3
32^3+46^3+82^3 = 88^3
17^3+40^3+86^3 = 89^3
10^3+60^3+80^3 = 90^3
25^3+38^3+87^3 = 90^3
45^3+60^3+75^3 = 90^3
58^3+59^3+69^3 = 90^3
6^3+72^3+74^3 = 92^3
54^3+60^3+74^3 = 92^3
32^3+54^3+85^3 = 93^3
15^3+50^3+90^3 = 95^3
19^3+53^3+90^3 = 96^3
48^3+64^3+80^3 = 96^3
45^3+69^3+79^3 = 97^3
11^3+66^3+88^3 = 99^3
16^3+68^3+88^3 = 100^3
35^3+70^3+85^3 = 100^3
. . .

 

[robert Ihnot]

However as a generalization from 3^2+4^2=5^2, it does not seem to stretch very far.

What I meant is that I know of no cases where four 4th powers equal a fourth power, let alone in consecutive integers.

Euler's Conjecture is that it took n nth powers to equal an an nth power. But it was shown in 1966--largely by accidential discovery that four fifth powers will do:

27^5+84^5+110^5+135^5 = 144^5.

 

[Char. Limit]

Euler's Conjecture is that it took n nth powers to equal an an nth power. But it was shown in 1966--largely by accidential discovery that four fifth powers will do:

27^5+84^5+110^5+135^5 = 144^5.

No it doesn't.

 

[bp_psy]

No it doesn't.

<br /> 27^5+84^5+110^5+133^5 = 144^5.<br />
http://www.wolframalpha.com/input/?i=%2827^5%2B84^5%2B110^5%2B133^5%29^%281%2F5%29

 

[Char. Limit]

Oh, now it does...

 

[TimGrebin]

The earlier confusion about the cube rooting, I think you have it wrong.

You have

a³ + b³ = c³

To cube root 'a', you are effectively multiplying it by a^1/3 , to do this you would have to also multiply b and c by a^1/3. Doing the same for b and c, you'd end up with an equation looking like this:

ab^1/3c^1/3 + ba^1/3c^1/3 = ca^1/3b^1/3

 

[Gregg]

The earlier confusion about the cube rooting, I think you have it wrong.

You have

a³ + b³ = c³

To cube root 'a', you are effectively multiplying it by a^1/3 , to do this you would have to also multiply b and c by a^1/3. Doing the same for b and c, you'd end up with an equation looking like this:

ab^1/3c^1/3 + ba^1/3c^1/3 = ca^1/3b^1/3

This isn't right
a^{1\over 3} = (-b^3+c^3)^{1\over 9}

 

[TimGrebin]

Good point.

 

[iamafigment]

(1^1/3)³ + (1^1/3)³ = (2^1/3)³

Why won't that work?

 

[Mentallic]

(1^1/3)³ + (1^1/3)³ = (2^1/3)³

Why won't that work?

This does work.

As I understand it, you are looking for values for a,b, and c where
a+b=c AND a^3 + b^3 = c^3

How about this?
a=cube rt 1
b=cube rt 1
c=cube rt 2

This doesn't work.

 

[disscombobila]

I have heard of this referred to as Pythagora's second theorem. It is said to be one of the most difficult and yet unsolved math problems known. It is stated, A cubed plus B cubed cannot equal C cubed and finding the proof for this has eluded the best mathematicians for centuries. According to lore, Pythagoras had the proof written down on a corner of a papyrus but some how the corner went missing. Props to anyone who can find it. I believe some think it may have to do with a Torus Knot.

 

[Char. Limit]

Already got a proof for a more general case - see Fermat's Last Theorem.

 

[Number Nine]

I have heard of this referred to as Pythagora's second theorem. It is said to be one of the most difficult and yet unsolved math problems known. It is stated, A cubed plus B cubed cannot equal C cubed and finding the proof for this has eluded the best mathematicians for centuries. According to lore, Pythagoras had the proof written down on a corner of a papyrus but some how the corner went missing. Props to anyone who can find it. I believe some think it may have to do with a Torus Knot.

As was said above, the problem has been solved (for quite some time). The lore you refer to involves Fermat, not Pythagoras.