Wikipedia:Reference desk/Archives/Mathematics/2010 January 22

= January 22 =

Weaker versions of Fermat's Last Theorem
Andrew Wiles proved that Fermat's Last Theorem is true, i.e. that for the equation $$a^n + b^n = c^n$$, there are no solutions where a, b, c and n are all natural numbers and n is greater than 2. What if we weaken the preconditions? If we allow at least one of a, b, and c to be any positive real, then it is trivially true that $$a^n + b^n = c^n$$ has solutions, because it reduces to checking whether $$(a^n + b^n)^{1/n}$$ is a positive real, and such a positive real always exists. But what if we say that a, b, and c have to be natural numbers, but n can be any positive real? What can we say about what solutions exist for what values of n? J I P  | Talk 07:04, 22 January 2010 (UTC)


 * For all natural numbers (a, b, c) such that c > a and c > b and c < a + b there exists a real number n such that $$a^n + b^n = c^n$$. I've discovered a lovely little proof of this, but this edit window is too small to contain it. Dragons flight (talk) 08:04, 22 January 2010 (UTC)
 * Let me try: ⊿. --Stephan Schulz (talk) 08:25, 22 January 2010 (UTC)
 * :-) Nice actually you don't need c83 but 64+74<84, so there is a solution with a=6, b=7, c=8, and n somewhere between 3 and 4.--RDBury (talk) 13:35, 22 January 2010 (UTC)


 * I don't have a proof but just thinking it through I'm sure there's an infinity of a solution for n: take e.g. the above and double all the numbers. The solution will be the same but you can also modify a, b and c to get a slightly different equation with a different solution. Rinse and repeat with ever bigger numbers. There should be $$\aleph_0$$ equations with solutions and so $$\aleph_0$$ values for n. But there are $$\aleph_1$$ real numbers in any interval, so an n picked at random will not be a solution. I don't know if you can say anything about the values for n other than this. -- JohnBlackburne wordsdeeds 13:49, 22 January 2010 (UTC)
 * Strictly speaking there are at least $$\aleph_1$$ real numbers in any interval. AndrewWTaylor (talk) 14:15, 22 January 2010 (UTC)


 * The homogeneity of the equation suggests an alternative proof to the one which (I assume) Dragons flight had in mind. Show that for any real u, v such that 0 < u < 1 and 0 < v < 1 there is a real w > 0 such that $$u^w + v^w = 1$$. Then take u = a/c and v = b/c. Gandalf61 (talk) 14:31, 22 January 2010 (UTC)
 * If w = 0 then $$u^w + v^w = 2$$. The limit as w→+∞ is 0. So by continuity there is a value of w between 0 and +∞ where the expression =1. There are only countably many choices for u and v so the number of possible values for w is countable. In particular Fermat's last theorem says w can't be an integer greater than 3. It probably wouldn't be hard to show the possible ws are dense in some interval though. So even though you can't have w=3 you could get w arbitrarily close to 3.--RDBury (talk) 18:04, 22 January 2010 (UTC)


 * A variation on these lines: for a given positive integer n it's not hard to find countably many solutions in positive integers x,y,z of the Diophantine equation x1/n+y1/n=z1/n. But can we characterize all solutions? In particular, is it necessary that x,y,z are perfect n-th powers? --84.220.119.131 (talk) 22:58, 22 January 2010 (UTC)

Bounds
Let us assume that a(t) and b(t) are, over ]0,∞[, both real valued and infinitely differentiable functions of the variable t. Furthermore, let us assume that the function c(t) is defined by c(t) = a(t) / b(t). Now, let us assume that the limit of c(t) as t tends towards positive infinity is k, where k is a positive real number. What can we say about a(t) and b(t)? What can we say about anything? Fly by Night (talk) 18:42, 22 January 2010 (UTC)
 * What sort of things do you want to say? We can say, for example, that b is eventually nonzero (that's the only constraint on b alone, though) and that a and b are eventually of the same sign. Algebraist 18:48, 22 January 2010 (UTC)
 * I wanted something more substantial than that. If we have two non-zero integers, say a and b, then we can form the number a/b. The integers cross the integers form a quotient space, namely the rational numbers: we say that (a,b) is in the same class as (c,d) if and only if ad = bc. Is their some kind of classification of pair of functions given a limit of their quotient? Fly by Night (talk) 19:05, 22 January 2010 (UTC)
 * There might be two equivalence relations. Define A(b, k) the set of functions which have limit k when divided by b, define B(a, k) the set of functions which have limit k when multiplied by a. My guess is that for every k, for every triplett of functions b, x, y it is true that: x in A(b, k) and y in A(b, k) => B(x, k) = B(y, k). Then B(A(b, k), k) could be an equivalence class for b. Maybe k is also irrelevant, so we get the class "all functions with similar limit behaviour as b". I don't know if that is actually true, but it sounds interesting and superficially plausible.  —Preceding unsigned comment added by 80.226.1.7 (talk) 15:45, 23 January 2010 (UTC)

Simplify by substituting t=1/x because an infinite t is more confusing than a zero x. Pick an arbitrary positive continuous function C(x) with C(0)=k. Pick an arbitrary positive continuous function B(x). Define A(x)=B(x)C(x). All you can say is that limx&rarr;0A(x)/B(x)=k. Bo Jacoby (talk) 22:25, 22 January 2010 (UTC).