Number theory challenge: find the flaw
http://kryptograff5.blogspot.com/2014/02/number-theory-challenge-find-flaw.html
Everyone thinks there ought to be an elementary proof of Fermat's last theorem. However, such attempts have all been in vain. There is always some Achilles' heel. So, what is there for most of us to do other than accept Andrew Wiles's proof on faith.
But wouldn't it be nice if an elementary proof turned up? That would in no way diminish Wiles's accomplishment, which doubtless has spurred whole areas of advancement in mathematics. On the other hand, an elementary proof is unlikely to contribute much to the development of mathematics.
Yet, everyone loves a puzzle! So, enjoy yourself and see if you can find the Achilles' heel in my little effort.
Concerning a noted result of number theory
1. Any three lengths can be arranged into a triangle.
2. So if a, b and (an + bn)1/n were all integers, then these three lengths would form a triangle.
3. For n = 2m+1 and m > 0, a, b and c ( = (an + bn)1/n ) form a non-right triangle.
4. Momentarily not worrying about whether n is even or odd, let n = 2t, we have
[(at)2 + (bt)2)]1/(2t) < [ (at)2 + (bt)2)]1/2
5. Or,
[an + bn]1/n <
[an + bn]1/2
6. And statement 3. is established.
7. A standard proof of the law of cosines places the longest side c, off the x axis.
8. We shall use this law, but also establish another relation by placing
c on the x axis and the altitude of the triangle ABC (the capitalized
letters represent the vertices) on the y axis. We call the origin 0, so
that C0 is the altitude, which we call h.
We say the length BO = r and the length
0A = c - r. Arbitrarily, a < b < c, with a =/= 0.
8a. That is, we have two right triangles joined by h, which is
perpendicular to c. Under side b on the x axis is base r and under side a
on the x axis is (c - r).
9. We obtain the simultaneous equation:
(c - r)2 + h2 = b2
r2 + h2 = a2
10. Which gives
c2 -2cr = b2 - a2
10a. Note that for statement 2. to hold, r must be rational.
11. Or
c2 = b2 - a2 + 2cr
12. We also have the cosine formula
c2 = b2 + a2 - 2abcos(c)
13. Combining
c2 = b2 - a2 + 2cr
c2 = b2 + a2 - 2abcos(c)
14. We obtain
c2 = b2 + cr - abcos(c)
15. Or
c(c - r) = b(b - acos(c))
16.
So that
c(c - r)/(b - acos(c) ) = b
16a. Note that this identity does not apply for n = 2, as cos(c) = 0 and c2 =/= b2 - a2 + 2cr.
17. Well
cn - bn = an
18. And
cn - cn(c - r)n/(b - acos(c))n = an
19. So
cn[1 - (c - r)n/(b - acos(c) )n] = an
19a. Note that as c - r and a are both positive, we require acos(c) > b/a.
19b. Check: let cos(c) = b/a.
19c. Then,
c2 = b2 + a2 - 2ab(b/a)
19d. Or, c2 = a2 - b2
which, by condition b > a, would make the right side negative (and c a complex number).
20. Anyway
c[1 - (c - r)n/(b - acos(c))n]1/n = b
21. Case i.
Suppose (c - r)/(b - acos(c)) = p/q where p and q are relatively prime integers.
In that case, we are done.
22. Case ii.
Suppose p/q = -j = |k|, where k and j are integers. (We are assuming n = 2m+1.)
23. So then
c(kn + 1)1/n = b
24. But
kn < kn + 1 < (k + 1)n
25. Or
k < (kn + 1)1/n < k + 1
26. Hence (kn + 1)1/n is no integer and can't be rational at all. So case ii is satisfied.
27. Now we are done.
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