Does the equation a³ + b³ = c³ have any geometrical significance?

[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.

$$27^5+84^5+110^5+133^5 = 144^5.$$
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.

 

[Loren Booda]

Search "Andrew Wiles."

 

[micromass]

To be fair, we don't need the full generality of Wiles' proof for the equation in the OP. The proof of the case $a^3+b^3=c^3$ can be found here: http://fermatslasttheorem.blogspot.be/2005/05/fermats-last-theorem-proof-for-n3.html

 

[iansinclair]

Hi guys this proof does exist beyond any reason of doubt and it proves itself.

a^n+b^n=c^n rings true as well. Obviously i cannot divulge much about this but WC was stabbing in the wrong direction as such.

as mentioned the proof proves itself.

The only thing WC got right was that it is a curve.

Hopefully i find the right place to publish one day but till then my safe is a good place as my Son and I had painstakingly worked our own logbook on this making it a mathematical equation that works and applies to modern maths.

Oh b t w you use fractions as well in the nth root as it is a great necessity.

Ian

 

[Diffy]

Hi guys this proof does exist beyond any reason of doubt and it proves itself.

a^n+b^n=c^n rings true as well. Obviously i cannot divulge much about this but WC was stabbing in the wrong direction as such.

as mentioned the proof proves itself.

The only thing WC got right was that it is a curve.

Hopefully i find the right place to publish one day but till then my safe is a good place as my Son and I had painstakingly worked our own logbook on this making it a mathematical equation that works and applies to modern maths.

Oh b t w you use fractions as well in the nth root as it is a great necessity.

Ian

LOL what? I can tell you are a mathematician when you say things like "the proof proves itself"

w t f does that even mean.. i don't even?

 

[micromass]

This thread is done.