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.
