Thread: Fermat's Last Theorem

i suspect some of the confusion here is arising because of the term "remainder". i believe the "remainder" obtained by the OP is the remainder of (x+b)ⁿ upon polynomial division by 2x+a.

this is *not* the same remainder obtained by integer division. for example, if x = 1, y = 3, z = 4, and n = 1 (so that a = 2, and b = 3), then the integer remainder is 0, because:

2x+a = 2+2 = 4
x+b = 1+3 = 4

and 4 divides 4.

it is in fact true that:

x+b = (1/2)(2x+a) + (b-(a/2)), but the quantity b-(a/2) is, in this case, 2, which is not 0 (nor even equivalent to 0 mod 2x+a = 4).

Richard1234 - It looks like your coefficients have gotten reversed - it is supposed to be 2x + a, not 2a + x. Hopefully that helps (or maybe that makes it worse for me!)

ebaines - You are correct about n=1. I never thought to go there. I will change my restriction on odd integers to be odd integers >= 3. This will prevent x+y = z since (x+y)^n > z^n for n>=3 if x^n + y^n = z^n. Allowing x+y=z seems to be what is letting you find the problems you found. Let me know.

Thank you both for your comments. I am quite new at this as you probably tell but enjoying it!

Thank you, Deveno. That helps tremendously and it gives me something to think about. I can see I am confusing integer division and polynomial division so it's definitely something for me to dig into and see where it takes me.

Originally Posted by tjs94080

Richard1234 - It looks like your coefficients have gotten reversed - it is supposed to be 2x + a, not 2a + x. Hopefully that helps (or maybe that makes it worse for me!)

Whoops, sorry about that. But even with the same variables, x = 5, a = 2, b = 4, n = 3, , which leaves a remainder of 9. And is still equal to 27.

Also weird is that 9 is not congruent to 27 (mod 12).

My derivation of the remainder being (b - (a/2))^n is contingent on choices of x,y,z,a,b,n such that x^n + y^n = z^n As your choices for those amounts don't conform to that restriction, it not surprising to me that 9<>27. Actually I just realized that given that Dr. Wiles already proved FLT true, you shouldn't ever be able to find a workable group of values where the actual remainder = (b - (a/2))^n.

My very limited and very distant exposure to abstract algebra (it's been a long time since college), I am unfamiliar with the congruence issues you raised. Could this be related to the differences Deveno alluded to that exist between integer division and polynomial division? As I am restricting my values of x,y,z,a,b,n to integers, I am not sure which additional properties of polynomial division exist here. The more I keep at this the more I am (re)learning!

I am going to rethink what I just said in my 4:46 pm post. Something seems to be unsettling about it.

I now see my problem with my statement 5. I understand the polynomial remainder of [b - (a/2)]^n is not required to be "the" remainder when integers are used. Thanks to everyone for helping me work through this. Besides, working on this problem off and on over the past 30+ years has been a good hobby.