# Fermat's Theorem

**URL:** <https://boards.straightdope.com/t/fermats-theorem/582989>\
**Category:** Factual Questions\
**Created:** [May 24, 2011, 5:58am UTC](https://boards.straightdope.com/t/fermats-theorem/582989 "2011-05-24T05:58:23Z")\
**Posts on this page:** 20\
**Page:** 7

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [May 27, 2011, 9:04pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/121 "2011-05-27T21:04:24Z")

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I think we all are still having a little trouble understanding exactly what your claim is.

Maybe try this for now, just to help us see what you’re saying: For each value of n from 2 through 10, tell us exactly what the corresponding claim is. [For now, just as we try to understand what sort of claim you are making, don’t try and give a single formula that covers all those cases at once; I mean, say “When n = 2, my claim is blah blah blah. When n = 3, my claim is blah blah blah”].

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**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\
**Post date:** [May 27, 2011, 9:43pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/122 "2011-05-27T21:43:56Z")

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**mla1** , is this what you’re saying:

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

has no solutions for n, x, and a[sub]i[/sub] all positive integers, for 1 \< p \< n-1.

So for example, you’re conjecture says that

x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 + a[sub]3[/sub]^5

has no solutions?

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<div class="post-metadata">

**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [May 27, 2011, 10:01pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/123 "2011-05-27T22:01:11Z")

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> [@ZenBeam](#):
>
> **mla1** , is this what you’re saying:
> 
> x^n = a[sub]1[/sub]^n + a[sub]2[/sub]^n + … + a[sub]n-p[/sub]^n
> 
> has no solutions for n, x, and a[sub]i[/sub] all positive integers, for 1 \< p \< n-1.
> 
> So for example, you’re conjecture says that
> 
> x^5 = a[sub]1[/sub]^5 + a[sub]2[/sub]^5 + a[sub]3[/sub]^5
> 
> has no solutions?

If so, that’s the special case of the Lander-Parkin-Selfridge conjecture mentioned at the bottom of the Wikipedia article on [Euler’s sum of powers conjecture](http://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture). So it seems at least no one’s ever disproven it…

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 28, 2011, 11:34am UTC](https://boards.straightdope.com/t/fermats-theorem/582989/124 "2011-05-28T11:34:36Z")

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x^n = a1^n + a2^n + … + an-p^n

That’s what i’m saying and p is and element of whole numbers.

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<div class="post-metadata">

**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [May 28, 2011, 4:38pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/125 "2011-05-28T16:38:15Z")

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What about it? For every n, there is a p which makes that equation solvable in integers? For every n and p that equation is solvable in integers? For every n, there is a p which makes that equation unsolvable in integers? You are not being clear enough; you are simply repeating the same formula we have trouble understanding what you are claiming about.

Again, maybe try this for now, just to help us see what you’re saying: For each value of n from 2 through 10, tell us exactly what the corresponding claim is. [For now, just as we try to understand what sort of claim you are making, don’t try and give a single formula that covers all those cases at once; I mean, say “When n = 2, my claim is blah blah blah. When n = 3, my claim is blah blah blah”.

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 1:49pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/126 "2011-05-29T13:49:00Z")

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n=2 p=0  
n=3 p=0  
n=4 p=1  
n=5 p=1

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**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\
**Post date:** [May 29, 2011, 2:36pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/127 "2011-05-29T14:36:27Z")

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> [@mla1](#):
>
> n=2 p=0  
> n=3 p=0  
> n=4 p=1  
> n=5 p=1

Seriously? If you are really a student, you understand that partial and incomplete responses are totally unacceptable.

What we are looking for is a SINGLE, COMPLETE statement of your claim.

So, for n= 2, p =0, what is the actual claim? That a^2 = b^2 + c^2 has solutions? Has no solutions? What?

For n = 3, p = 0, what is the actual claim? That a^3 = b^3 + c^3 + d^3 has solutions? No solutions? Solutions only under other conditions?

Likewise n = 4, p =1 and n=5,p=1.

What is your actual claim? Be specific and be complete. A single sentence won’t cut it. Be thorough. Completely explain what you are trying to explain. Use examples. It should tell you a lot that nobody, including working mathematicians, can figure out what you are trying to claim. It means your statement is, at best, vague.

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**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 2:43pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/128 "2011-05-29T14:43:08Z")

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I don’t understand i’m only 16 what part of the claim must i explain indistinguishable wanted when n= this p= that.

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 2:47pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/129 "2011-05-29T14:47:09Z")

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do you mean x^2 = a1^2 + a2-0^2  
x^3 = a1^3 + a2^3 +a3-0^3  
x^4 = a1^4 + a2^4 + a3-1^4  
x^5 = a1^5 +a2^5 + a3^5 +a4-1^5  
Therefore x^n = a1^n +a2^n +a3^n + . . . an-p^n

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 2:54pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/130 "2011-05-29T14:54:28Z")

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Antibob if you can not understand my claim please ask me which part you want to explain. When you do please take into account that I’m 16 and do not understand many (if any) complex functions. There are things you know that I cannot grasp as I have not been taught them. Also be specific about what part of the conjecture and how i should lay it out so that in future i don’t cause such great confusion.

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**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\
**Post date:** [May 29, 2011, 3:13pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/131 "2011-05-29T15:13:57Z")

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So you’re saying that when you say:

n=2 p=0

what you mean is:

x^2 = a1^2 + a2-0^2 .

And when you say:

n=3 p=0

what you mean is:

x^3 = a1^3 + a2^3 +a3-0^3 .

And when you say:

n=4 p=1

what you mean is:

x^4 = a1^4 + a2^4 + a3-1^4 .

So you’re saying that when you say:

n=5 p=1

what you mean is:

x^5 = a1^5 +a2^5 + a3^5 +a4-1^5 .

O.K., let’s start with the first of these claims. What does it mean to say that

x^2 = a1^2 + a2-0^2 ?

First of all, what’s in subscript and what’s in superscript here? If you can’t use standard subscript and superscript notation, you’re going to have to explain what’s going on for each equation. I presume that x^2 means “x squared”. I presume that a1 means “a with a 1 subscript”. I presume that a1 means “a with a 1 superscript” and a1^2 means that the variable a1 has been squared. That’s not obvious. Because you aren’t using superscripts and subscripts, it could mean that 1 is squared and then that is the subscript of a. I can’t figure out what a2-1^0 means at all. Is “-0” in the subscript or the superscript? Are you subtracting 0 from 2 and letting that be the subscript to a? Or is 0 being subtracted from 2? Or is 0 being squared?

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 3:33pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/132 "2011-05-29T15:33:32Z")

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how do you subscript and supercript

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<div class="post-metadata">

**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\
**Post date:** [May 29, 2011, 3:44pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/133 "2011-05-29T15:44:11Z")

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Will someone explain to mla1 how superscripting and subscripting are done on the SDMB? I don’t understand that myself.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [May 29, 2011, 3:48pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/134 "2011-05-29T15:48:22Z")

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> [@mla1](#):
>
> how do you subscript and supercript

[noparse]x[sub]2[/sub][/noparse] yields x[sub]2[/sub].

[noparse]x[sup]2[/sup][/noparse] yields x[sup]2[/sup].

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 4:04pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/135 "2011-05-29T16:04:47Z")

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x [sup]2]sup] = a[sup]2[sup][sub]1-0[sub]  
x[sup]3[sup] = a[sup]3[sup][sub]1[sub] + a[sup]3[sup][sub]2 + a[sup]3[sup][sub]3-0[sub]  
x[sup]4[sup] = a[sup]4[sup][sub]1[sub] + a[sup]4[sup][sub]2[sub] + a[sup]4[sup][sub]4-1[sub]  
x[sup]5[sup] = a[sup]5[sup][sub]1[sub] + a[sup]5[sup][sub]2[sub] + a[sup]5[sup][sub]3[sub] +a[sup]5[sup][sub]5-1][sub]  
x[sup]n[sup] = a[sup]n[sup][sub]1[sub] + a[sup]n[sup][sub]2[sub] + a[sup]n[sup][sub]3[sub] + . . . a[sup]n[sup][sub]n-p[sub]

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 4:11pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/136 "2011-05-29T16:11:21Z")

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x [sup]2[/sup] = a[sup]2[/sup][sub]1[/sub] + a[sup]2[/sup][sub]2-0[/sub]  
x[sup]3[/sup] = a[sup]3[/sup][sub]1[/sub] + a[sup]3[/sup][sub]2[/sub] + a[sup]3[/sup][sub]3-0[/sub]  
x[sup]4[/sup] = a[sup]4[/sup][sub]1[/sub] + a[sup]4[/sup][sub]2[/sub] + a[sup]4[/sup][sub]4-1[/sub]  
x[sup]5[/sup] = a[sup]5[/sup][sub]1[/sub] + a[sup]5[/sup][sub]2[/sub] + a[sup]5[/sup][sub]3[/sub] +a[sup]5[/sup][sub]5-1[/sub]  
x[sup]n[/sup] = a[sup]n[/sup][sub]1[/sub] + a[sup]n[/sup][sub]2[/sub] + a[sup]n[/sup][sub]3[/sub] + . . . a[sup]n[/sup][sub]n-p[/sub]

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<div class="post-metadata">

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [May 29, 2011, 4:18pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/137 "2011-05-29T16:18:16Z")

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> [@mla1](#):
>
> x [sup]2[/sup] = a[sup]2[/sup][sub]1[/sub] + a[sup]2[/sup][sub]2-0[/sub]  
> x[sup]3[/sup] = a[sup]3[/sup][sub]1[/sub] + a[sup]3[/sup][sub]2[/sub] + a[sup]3[/sup][sub]3-0[/sub]  
> x[sup]4[/sup] = a[sup]4[/sup][sub]1[/sub] + a[sup]4[/sup][sub]2[/sub] + a[sup]4[/sup][sub]4-1[/sub]  
> x[sup]5[/sup] = a[sup]5[/sup][sub]1[/sub] + a[sup]5[/sup][sub]2[/sub] + a[sup]5[/sup][sub]3[/sub] +a[sup]5[/sup][sub]5-1[/sub]  
> x[sup]n[/sup] = a[sup]n[/sup][sub]1[/sub] + a[sup]n[/sup][sub]2[/sub] + a[sup]n[/sup][sub]3[/sub] + . . . a[sup]n[/sup][sub]n-p[/sub]

Are you asserting that there are integer solutions? That there aren’t? Complete sentences are vital here. A very eminent mathematician once told me that ideas are only valuable if you can communicate them clearly, and I have never been given any reason to doubt him.

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 4:46pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/138 "2011-05-29T16:46:54Z")

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Ltrafilter that sounds like a very wise man and the type of person i would love to meet. wrote that out to demonstrate the conjecture. I did now attach anything to make sure that is was understood. There are solutions. Assuming that p is a whole number and x and a are part the natural number system.

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<div class="post-metadata">

**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\
**Post date:** [May 29, 2011, 5:14pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/139 "2011-05-29T17:14:38Z")

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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 no solutions for n, p, x, and a[sub]i[/sub] all positive integers, n \>= 4, for 1 \< p \< n-1  
and has no solutions for n, p, x, and a[sub]i[/sub] all positive integers, n = 3, for 0 \< p \< n-1 (i.e., p = 1 has no solutions either).  
I guess he’s including n = 2 cases as well, but there aren’t any values of p for which there aren’t solutions.

So

x^2 = a[sub]1[/sub]^2 + a[sub]2[/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[/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]3[/sub]^4 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]4[/sub]^5 has solutions.

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 has solutions.

etc.

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<div class="post-metadata">

**Author:** ![mla1](https://avatars.discourse-cdn.com/v4/letter/m/c0e974/32.png) [@mla1](https://boards.straightdope.com/u/mla1)\
**Post date:** [May 29, 2011, 5:31pm UTC](https://boards.straightdope.com/t/fermats-theorem/582989/140 "2011-05-29T17:31:24Z")

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Zedbeam you so close to understanding what I’m saying. But you put something in there that will make the equation false for some expoenents. What I’m actually saying is.

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

has solutions as long as n is a whole number and as long as p is a whole number  
and has solutions for n, p, x, and all positive integers, n = 0, for 0 \< p (i.e., p = 1 has solutions as long as n is a particular number).

In case you people in university forgot whole number are 0;1;2;3;4;5 . . .

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