Friday, February 14, 2014

Number theory challenge: find the flaw

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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