**Concerning a noted result of number theory**

1. Any three lengths can be arranged into a triangle.

2. So if a, b and (a

^{n}+ b

^{n})

^{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 ( = (a

^{n}+ b

^{n})

^{1/n}) form a non-right triangle.

4. Momentarily not worrying about whether n is even or odd, let n = 2t, we have

[(a

^{t})

^{2}+ (b

^{t})

^{2})

^{]1/(2t)}< [ (a

^{t})

^{2}+ (b

^{t})

^{2})

^{]1/2}

5. Or,

[a

^{n}+ b

^{n}]

^{1/n}< [a

^{n}+ b

^{n}]

^{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}+ h

^{2}= b

^{2}

r

^{2}+ h

^{2}= a

^{2}

10. Which gives

c

^{2}-2cr = b

^{2}- a

^{2}

10a. Note that for statement 2. to hold, r must be rational.

11. Or

c

^{2}= b

^{2}- a

^{2}+ 2cr

12. We also have the cosine formula

c

^{2}= b

^{2}+ a

^{2}- 2abcos(c)

13. Combining

c

^{2}= b

^{2}- a

^{2}+ 2cr

c

^{2}= b

^{2}+ a

^{2}- 2abcos(c)

14. We obtain

c

^{2}= b

^{2}+ 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 c

^{2}=/= b

^{2}- a

^{2}+ 2cr.

17. Well

c

^{n}- b

^{n}= a

^{n}

18. And

c

^{n}- c

^{n}(c - r)

^{n}/(b - acos(c))

^{n}= a

^{n}

19. So

c

^{n}[1 - (c - r)

^{n}/(b - acos(c) )

^{n}] = a

^{n}

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,

c

^{2}= b

^{2}+ a

^{2}- 2ab(b/a)

19d. Or, c

^{2}= a

^{2}- b

^{2}

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(k

^{n}+ 1)

^{1/n}= b

24. But

k

^{n}< k

^{n}+ 1 < (k + 1)

^{n}

25. Or

k < (k

^{n}+ 1)

^{1/n}< k + 1

26. Hence (k

^{n}+ 1)

^{1/n}is no integer and can't be rational at all. So case ii is satisfied.

27. Now we are done.

## No comments:

Post a Comment