Re: ---- ----- ---- An assertion in number theory

On Jan 17, 4:46 pm, Tonic...@xxxxxxxxx wrote:
On Jan 17, 9:24 pm, Deep <deepk...@xxxxxxxxx> wrote:





On Jan 17, 1:40 pm, Tonic...@xxxxxxxxx wrote:

On Jan 17, 7:54 pm, Deep <deepk...@xxxxxxxxx> wrote:

On Jan 17, 12:24 pm, Tonic...@xxxxxxxxx wrote:

On Jan 17, 6:59 pm, Deep <deepk...@xxxxxxxxx> wrote:

Consider the following two equation under the given conditions.

x^2k + y^2k = z^2k                    (1)

x^k + y^k = z^k                         (2)

Conditions: x, y, z are relatively prime positive integers, y is even
and prime k > 3.

Assertion: Set of x, y, z for which (1) is impossible will also be
impossible for (2).

My proof: Parametric solutions of (1) are given as:

x = z(coskD)^(1/k)     (3)          y = z(sin kD)^(1/k)       (4)

where 0 < D < pi/2

Verification: x^2k + y^2k = z^2k[ (coskD)^2 + (sin kD)^2] = z^2k

From (3) and (4) one gets (5)

x^k + y^k = z^k[ cok kD + sin kD]        (5)

Since by FLT (1) is impossible for integer x, y, z

Therefore, (6) is also impossible for integer x, y, z.

x^k + y^k = z^k                 (6)

Any helpful comments about the proof will be appreciated.

*************************************************************

If I didn't misunderstand something, I think FLT's proof made all the
above trivial.

Regards
Tonio- Hide quoted text -

- Show quoted text -

---- You didnot misunderstand anything. FLT has been proved without
any doubt.
This is just another way to look at it. Pl comment upon the
correctness of my approach.

*******************************************************

Ok...then I think cos(kD)^(1/k) , sin(kD)^(1/k) are not integers in
general....in fact, I think that there are not integers for ANY value
of k and D, so then your approach doesn't really meet the requirement
of x,y,z integers.

Regards
Tonio- Hide quoted text -

- Show quoted text -

---- Just kindly look differently. All I want to say that given (1)
has no integer solutions then (2) also cannot have any integer
solutions. The reverse is obvious.
Please let me know your arguments if you donot agree with my
assertion. Digression: Why are you so sure that (sin kD)^1/k and (cos kD)^1/k cannot be integers? Can you kindly offer a proof?

**********************************************************

Let's see: for any value 0 < D < Pi/2 and any integer k, both
cos(kD) and sin(kD) are numbers in [-1,1] ==> since the k-th root of a
rational number is a rational number iff it is an exact root (the k-th
root of an irrational number is, of course, irrational), it'd have to
be that cos(kD), sin(kD) are both exact k-th roots in order to be
rational numbers, and even then they wouldn't be integers unless they
both equal +/- 1, which of course cannot be for the same values of k
and D.

I hope I didn't miss anything above, and in any case, I think it is
you who would have to prove that your numbers x = z*cos(kD)^1/k  and
y = z*sin(kD)^1/k  are integers...

About what you ask for me kindly to look differently: I know you
wanted to prove something hasn't integers values, but during the proof
you used numbers that, apparently, are not integers and thus the proof
is, imo flawed.

Regards
Tonio- Hide quoted text -

- Show quoted text -

---- Thanks again. I am restating the assertion.
Assertion: If x, y, z are the real roots of (1) then (2) cannot
be satisfied.

x^(2k) + y^(2k) = z^(2k) (1)

x^k + y^k = z^k (2)

If you donot agree with the assertion I will offer additionaal
explanation if you ask for.
If you agree with the assertion I will feel happy.
Kindly respond.
.

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