Fermat's Theorem

Factual Questions
141

Ok, I’ll play.

A complete statement of a conjecture means that in a single post you give all the details necessary so that a reasonably informed person (say with a modicum of mathematical knowledge) can read that one, single post and understand your conjecture.

From the several posts requesting clarification, it is obvious you have not yet met this standard.

For example, let’s say that all I wrote was:

x^n + y^n = z^n

So far, this is YOUR style of posting. You give an equation but don’t explain your terms and simply let us flail about trying to understand how you are using the terms.

A complete statement would be:

The equation x^n + y^n = z^n, where x, y, z, and n are whole numbers, has no solutions for n > 2.

Your last post states that the equation has solutions for p > 0. Are there any other restrictions on p? By construction, obviously p < n, so we have 0 < p < n. But you also use p = 0 when n = 2 and n = 3. So, which is it? p >= 0? Or p > 0? Also, clearly, if n > 2, then p = n-2 will have no solutions (that’s Fermat’s last theorem, after all).

ZenBeam (and you should really try to get usernames correct - it’s just basic etiquette) has done an admirable job of trying to “interpret” your posts to make them understandable to the rest of us but still gets it wrong (at least according to you).

So, my suggestion is to take one of ZenBeam’s posts and simple copy/paste. Edit it so that there are no longer incorrect statements. That would make for a very complete set of statements that we could use to understand your basic point.

142

n=2 p=0
n=1 p=0
Ok. I’ll do that. gove me time to take what he said and make it correct. I’l post it later today. Zedbeam thak you for attempting to get it.

143

I’m fairly certain what mla1 is trying to say is:

x^n = a[sub]1[/sub]^n + a[sub]2[/sub]^n + … + a[sub]n-p[/sub]^n

has solutions for n, p, x, and all positive integers, n >= 0, for 0=< p < n-1
if n=2 p=0
if n=1 p=0
if n=3 p=0
(i.e., p = 1 has no solutions either).

So

x^2 = a[sub]1[/sub]^2 + a[sub]2-0[/sub]^2 has solutions

x^3 = a[sub]1[/sub]^3 + a[sub]2[/sub]^3 does not have solutions.
x^3 = a[sub]1[/sub]^3 + a[sub]2[/sub]^3 + a[sub]3-0[/sub]^3 has solutions.

x^4 = a[sub]1[/sub]^4 + a[sub]2[/sub]^4 does not have solutions.
x^4 = a[sub]1[/sub]^4 + a[sub]2[/sub]^4 + a[sub]4-1[/sub]^4 has solutions.
x^4 = a[sub]1[/sub]^4 + a[sub]2[/sub]^4 + a[sub]3[/sub]^4 + a[sub]4-0[/sub] has solutions

x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 does not have solutions.
x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 + a[sub]3[/sub]^5 does not have solutions.
x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 + a[sub]3[/sub]^5 + a[sub]5-1[/sub]^5 has solutions.
x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 + a[sub]3[/sub]^5 + a[sub]4[/sub]^5 + a[sub]5-0[/sub]

x^6 = a[sub]1[/sub]^6 + a[sub]2[/sub]^6 does not have solutions.
x^6 = a[sub]1[/sub]^6 + a[sub]2[/sub]^6 + a[sub]3[/sub]^6 does not have solutions.
x^6 = a[sub]1[/sub]^6 + a[sub]2[/sub]^6 + a[sub]3[/sub]^6 + a[sub]4[/sub]^6 does not have solutions.
x^6 = a[sub]1[/sub]^6 + a[sub]2[/sub]^6 + a[sub]3[/sub]^6 + a[sub]4[/sub]^6 + a[sub]5[/sub]^6 does not have solutions.
x^6= a[sub]1[/sub]^6 + a[sub]2[/sub]^6 + a[sub]3[/sub]^6 + a[sub]4[/sub]^6 + a[sub]5[/sub]^6 + a[sub]6-0[/sub] has solutions.

144

Ok, I think I see where you are going now. I can see 2 possible ways this can go.

  1. Simple existence statement. There exists at least one p, 0 <=p < n-1, such that the equation:

     b[sup]n[/sup] = a[sub]1[/sub][sup]n[/sup] + a[sub]2[/sub][sup]n[/sup] + ... + a[sub]n-p[/sub][sup]n[/sup]

has solutions.

  1. Beyond existence - for (1) above, the values of p for which there are solutions to the equation are ???
145

the first thing you said is right.

146

You included the n = 1 and n = 0 cases. n = 1 is trivial:

x^1 = a[sub]1[/sub]^1

But how does the n = 0 case work? The right hand side has no terms:

x^0 =

I guess you could say it’s 0, but then you’d get

x^0 = 0

which doesn’t have any solutions. You’d have to let p = -1 to get an a[sub]1[/sub] term.

147

yes it does x^0=a^0

148

That has n = 0 and p = 1, which doesn’t jibe with Great Antibob’s statement that you agreed with.

149

no he said pis equal to or greater than 0

150

When I was a kid…

… a six killed my dad.

151

It requires p = -1, not p = +1. The last term is a[sub]n-p[/sub]^n, so n-p has to be at least 1. Since n = 0, that requires p = -1.

ETA: Either way, the n = 0 and n = 1 cases are boring. Just wanted to clarify what I was saying.

152

And less than or equal to n - 1.

153

Quoth Bryan Eckers:

Well, you know, six is afraid of seven.
Because 7 8 9.

154

Zedbeam x^0 =a[sub]1-0[/sub]^0

155

But a[sub]1-0[/sub] is a[sub]n-p[/sub], so that would be n = 1, p = 0, not n = 0, p = 1.

Or, to say it a different way, n-p = 1, and n = 0, so you have p = -1.

And it’s ZenBeam, as Great Antibob pointed out several posts ago.

156

never mind you know what Antibob wrote? can you prove it

157

Prove what? That when n-p = 1 and n = 0, then p = -1?

158

no the theorem

159

n-p does not always =1 it can =2 it can =1

160

If you guys are willing to try and prove it I will work on patterns and things that will make the proof easier but I need at least one person to try have a go. If you still don’t understand the formula don’t be afraid to ask and I’ll do my best to answer.