Power Laws

shahar

shahar

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In 1825 A. L. prove that:
a5+b5!=c5
and L. Euler prove in 1753 that:
a3+b3!=c3

Hence
a15+15!=c15
Doesn't exist. because 3x5 (???)
Why the conclusion is true?
Why the power of 15 dependent on the power of 3 and 5. How does it can be proven from the power of 3 and power of 5 that power of 15 doesn't exist.
a, b and c are Natural.
 
No, they did not prove that. You are not stating clearly what they proved. They proved that there are no integers, a, b, and c, such that \(\displaystyle a^3+ b^3= c^3\) and no integers such that \(\displaystyle a^5+ b^5= c^5\).

It does NOT follow from that that there are no integers such that \(\displaystyle a^{15}+ b^{15}= c^{15}\). You cannot combine equations like that.
 
HallsofIvy said:
No, they did not prove that. You are not stating clearly what they proved. They proved that there are no integers, a, b, and c, such that \(\displaystyle a^3+ b^3= c^3\) and no integers such that \(\displaystyle a^5+ b^5= c^5\).

It does NOT follow from that that there are no integers such that \(\displaystyle a^{15}+ b^{15}= c^{15}\). You cannot combine equations like that.
Click to expand...
Yes! It has been proven that for integer numbers

`a^N + b^N \ne c^N` when `N \gt 2`

However,

`10^3 + 9^3 = 12^3 + 1^3` ...... my flabber was gusted when I first saw this identity
 
The sign != means no equal to.
and the presentation (that I can give a link [Hebrew Presentation) say comes because of power laws) the conclusion get from the power of 3 and 5. If the two inequalities not give that so you mean that is flaw in the presentaion?
(From the Technion Institute)
 
shahar said:
a5+b5!=c5
a3+b3!=c3

Hence
a15+15!=c15
Doesn't exist. because 3x5 (???)
Why the conclusion is true?
Why the power of 15 dependent on the power of 3 and 5. How does it can be proven from the power of 3 and power of 5 that power of 15 doesn't exist.
a, b and c are Natural.
Click to expand...
You are being careless in your typing, but what you appear to be saying is that this paper says that from the facts that there are no natural numbers that satisfy \(a^3+b^3=c^3\) or \(a^5+b^5=c^5\), we can conclude that there are no natural numbers such that \(a^{15}+b^{15}=c^{15}\).

Have you tried proving this by a simple contradiction?

Suppose that \(a^{15}+b^{15}=c^{15}\). Can you then find natural numbers x, y, and z such that \(x^3+y^3=z^3\)?
 
OK. I have a mistake. Somebody fix me, it go like this; (sorry of the harassment):
If a15+b15=c15 so (a5)3+(b5)3=(c5)3
and
a5, b5 and c5 exist
hence - the power of 3 exist.
Sorry, about the harassment
 
?? Yes, they exist. You could even argue that a, b, and c exist. It does not follow that they are integers nor that the same a, b, and c satisfy all of those equations. I really do not understand the point you are trying to make.
 
HallsofIvy said:
?? Yes, they exist. You could even argue that x, y, and z exist. It does not follow that they are integers or that the same x, y, and z satisfy all of those equations. I really do not understand the point you are trying to make.
Click to expand...
I don't understand what you want to say.
A^15 + B^15 ≠ C^15
came from the 2 other inequations or not?
 
shahar said:
I don't understand what you want to say. a^15 + b^15 ≠ c^15 came from the 2 other inequations or not?
Click to expand...
No! You said:

shahar said:
OK. I have a mistake. Somebody fix me, it go like this; (sorry of the harassment):
If a15+b15=c15 so (a5)3+(b5)3=(c5)3
and
a5, b5 and c5 exist hence - (the power of 3 exist→) A3 + B3 = C5.
Sorry, about the harassment
Click to expand...
So a3 + b3 = c3

came from

A15+B15=C15 .........→ ......... X3+Y3=Z3

where

X = a5........ & ........ Y = b5 ........ & ......... Z = c5

also

A15+B15=C15 ......... → ......... U5+V5 = W5

where

U = a3........ & ........ V = b3 ........ & ........ W = b3
 
Someone tell that I can do the original question by using the
Legendre's formula
in math theory. This is true?
 
@lex

Shahar is a frequent contributor and often asks questions about things he has read in articles written in Hebrew. Moreover, his English Is, to put it kindly, tenuous.

If I had to venture a guess, he was reading an article on the history of Fermat’s Last Theorem, and the article said something like

Given Abel’s and Euler’s results, it is possible to prove [MATH]a^x + b^x = c^x[/MATH] has no solution in integers if x is a multiple of 3 or 5.

He wondered why that was true.

So I think you and Dr. P have identified the question and provided the answer. Needless to say, the original post was not particularly well crafted.
 
I restate my question:
By the two theorems of 2 inequalities that say the power 3 and power 5 give the inequality of power 15. Why?
 
shahar said:
I restate my question:
By the two theorems of 2 inequalities that say the power 3 and power 5 give the inequality of power 15. Why?
Click to expand...
You still do not state the proposition correctly.

Furthermore, you need only one of those theorems to reach your result. Having both means that you can get there two different ways.

Consider the proposition below as a proven theorem.

[MATH]a,\ b \in Z^+ \text { and } i = 3 \text { or } 5 \implies \not \exists \text { positive integer } c \text { such that } c^i = a^i+ b^i.[/MATH]
From that we can prove by contradiction

[MATH]u, \ v, \text { and } k \in \mathbb Z^+ \text { and } i = 3 \text { or } 5 \implies \not \exists \text { positive integer } w \text { such that } w^{ki} = u^{ki} + v^{ki}.[/MATH]
Here is the proof, which basically lex already gave.

[MATH]\text {ASSUME, for purposes of contradiction, that } \exists \text { positive integers } u, \ v, \ w, \text { and } k \text { such that } u^{ki} + v^{ki} = w^{ki}.[/MATH]
[MATH]\therefore u^k = a, \ v^k = b, \text { and } c = w^k \implies a, \ b, \text { and } c \in \mathbb Z^+.[/MATH]
[MATH]\text {And } u^{ki} = a^i, \ v^{ki} = b^i, \text { and } w^{ki} = c^i.[/MATH]
[MATH]\therefore a^i + b^i = c^i, \text { which contradicts a previously proven theorem.}[/MATH]
[MATH]\text {Therefore, the assumption is false, and its opposite is true.}[/MATH]
 
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