Re: Fermat's Last Theorem

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**tjs94080** 5) For (4) to be true, the remainder of [(x+b)^n] / (2x +a) must be 0. It can be shown algebraically that the remainder is [b - (a/2)]^n, so this implies b = a/2

Any insights here would be appreciated.

How do you know this is true?

Re: Fermat's Last Theorem

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Originally Posted by

**tjs94080** I have been playing around with this off and on for several years.

Unless I did my algebra wrong, I am now coming up with the following "simple" proof. Anyone see an error here?

1) It can be shown algebraically that (x + y) divides (x^n + y^n) for odd integer n

2) Assume integers x, y, z ( x< y < z) exist such that x^n + y^n = z^n for odd interger n

3) Assume y = x + a and z = x + b. Since x < y < z, b > a > 0

4) From (1) and using (2) and (3) to substitute for y and z , (2x + a) divides (x + b)^n

5) For (4) to be true, the remainder of [(x+b)^n] / (2x +a) must be 0. It can be shown algebraically that the remainder is [b - (a/2)]^n, so this implies b = a/2

6) (5) contradicts (3) that b > a. This implies (2) cannot be true.

Any insights here would be appreciated.

What about for even integer values of n? Fermat's Last Theorem states that there are not any positive integer values of n > 2 such that x^n + y^n = z^n, not just odd integers...

Re: Fermat's Last Theorem

Prove It is right; Fermat's last theorem is for any integer n.

Although I would be quite impressed if you proved it for odd integer n with such a short proof. However part (5) in your solution appears flawed right now.

Re: Fermat's Last Theorem

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**richard1234** Prove It is right; Fermat's last theorem is for any integer n.

Although I would be quite impressed if you proved it for odd integer n with such a short proof. However part (5) in your solution appears flawed right now.

I would be equally impressed. I wonder if this is the sort of method Fermat himself claims he was following when he "proved" his own theorem...

Re: Fermat's Last Theorem

Again, assuming my arithmetic and algebra are correct, you should be able to see the pattern using long division of (x+b)^3 by (2x+a), (x+b)^5 by (2x +a); and (x+b)^7 by (2x+a). You will need to first expand (x+b)^3,5,7 and divide long hand. The final remainder simplifies to (b - (a/2))^n

Re: Fermat's Last Theorem

Also regarding even powers for n, it seems to be fairly well established in the math community that the FLT does not hold for even powers so I only honed in on a proof for odd powers.

Re: Fermat's Last Theorem

Yes, but even if the remainder is $\displaystyle (b-2a)^n$, can you claim that $\displaystyle b-2a = 0$? Maybe the remainder is something congruent to 0 (mod 2a + x). I'll have to take a look.

Re: Fermat's Last Theorem

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Originally Posted by

**tjs94080** Also regarding even powers for n, it seems to be fairly well established in the math community that the FLT does not hold for even powers so I only honed in on a proof for odd powers.

Actually, it has been proven by Wiles that Fermat's Last Theorem indeed does hold for all powers of n > 2. And the statement of the theorem does say "There are not any integers greater than 2 such that x^n + y^n = z^n". That definitely includes the even powers.

I suggest you watch this video if you want an idea of how he did it. Fermat's Last Theorem (Complete) - YouTube

Re: Fermat's Last Theorem

Prove It -Of course you are correct. Wiles' proof was all inclusive though definintely not simple. However, the literature (articles, lectures, books, etc. ) in existence before his proof overwhelmingly stated that merely showing a proof for all prime integer powers > 2 (which are all odd) would be adequate. I started my search many years before Wiles' proof and have only looked at the odd integer powers.

Richard1234 - I think we have a problem with all high school algebra courses if we can't take the nth root of both sides of a simple polynomial equation with integer coefficients and maintain equality. Given n will be odd, we don't even have a parity issue (+/-) that exists for even powers. However, given that the product of the n identical factors is 0, I believe the zero-product property requires at least one of the factors (or in this case all of the identical factors) be zero.

Thanks for the comments.

Re: Fermat's Last Theorem

Yeah, you might be right. I think it has to do with the fact that the long division method assumes that you find the most number of times (something) goes into 2a+x, thereby making the remainder as small as possible.

I was thinking that $\displaystyle (b-2a)^n \equiv 0 (\mod 2a+x)$ didn't necessarily imply $\displaystyle (b-2a)^n = 0$. You may want a second or third opinion. First, just make sure that your division is correct and that it holds for all n.

Re: Fermat's Last Theorem

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**tjs94080** the remainder of [(x+b)^n] / (2x +a) must be 0. It can be shown algebraically that the remainder is [b - (a/2)]^n, so this implies b = a/2

I think this step is incorrect. The remainder of [(x+b)^n] / (2x +a) is not [b - (a/2)]^n. I simply tried a few values of x, a, b, and n, and found it doesn't work. For example if x=4, a = 1, b= 3, n = 3 you get [(x+b)^n] / (2x +a) = 343/9, which has a remainder of 1, not a remainder of (3-1/2)^3 = 15.625.

Re: Fermat's Last Theorem

Thanks for the comment, ebaines.

However, I believe you have misunderstood my statement. I am noting that in order for (2x +a) to divide (x+b)^n, the ultimate remainder is this division must be zero or (2x+a) is not dividing (x+b)^n. You, in essence, showed that if 2b does not equal a, the remainder is not zero. I agree with you. The only way for the remainder to be zero is to have 2b = a. That is a necessary condition to make the remainder zero. Given this necessary condition, then b <= a (trivial case of both being zero) but it will not be possible for b <= a and still have Statements 2 and 3 be true where b > a because z > y. That is the heart of my indirect proof.

I have determined the general form of the quotient when dividing (x+b)^n by (2x+a) using long division (it is not pretty!). Algebraically, the remainder works out to be [b - (a/2)]^n. I am working on getting this information (including all details on the divisions in statements 1, 4, and 5) from paper to electronic form for a proper release and review.

Regards.

Re: Fermat's Last Theorem

Assume $\displaystyle a$ is even (otherwise $\displaystyle (b - \frac{a}{2})^n $ would not be an integer). However this implies x and y have the same parity, but whatever. Also, assume n is odd (since you want to prove for odd n).

Try x = 5, a = 2, b = 4, n = 3. Then $\displaystyle \frac{(x+b)^n}{2a+x} = \frac{(5+4)^3}{2(2)+5}$, the remainder is zero. But $\displaystyle (b - \frac{a}{2})^n = (4 - \frac{2}{2})^3 = 27$. Clearly, 0 does not equal 27.

What I've been trying to say all along is that you can't automatically claim $\displaystyle b - \frac{a}{2} = 0$ because the remainder *could be something congruent to* 0 (mod 2a + x), or 0 (mod 9). 27 ≡ 0 (mod 9).

Re: Fermat's Last Theorem

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Originally Posted by

**tjs94080** I have determined the general form of the quotient when dividing (x+b)^n by (2x+a) using long division (it is not pretty!). Algebraically, the remainder works out to be [b - (a/2)]^n. I am working on getting this information (including all details on the divisions in statements 1, 4, and 5) from paper to electronic form for a proper release and review.

OK - looking forward to it, because I've tried a lot of values and it doesn't seem to work . It certainly doesn't work for the trivial case of n=1.