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Topic: ---- --- --- --- validity of an equation
Replies: 5   Last Post: Mar 14, 2013 6:20 PM

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 Ulrich D i e z Posts: 34 Registered: 1/2/10
Re: ---- --- --- --- validity of an equation
Posted: Mar 14, 2013 6:20 PM

Deep wrote:

> Consider the following equation under the given conditions
>
> y^(1/2) = (x^10 - z^ 5)^(1/5 ) (1)
>
> Conditions: y, x, z are co prime integers each > 5, 2|y.
>
> Question: can (1) have any solution ?
>
> Any helpful comment will be appreciated

(English is not my first language. Hints about errors in
matters of speech/spelling/grammar etc will be appreciated.)

2|y means: An integer Y exists with y = 2Y :

(2Y)^(1/2) = (x^10 - z^5)^(1/5) (1')

Raising both "sides" of the equation to the power of 5
yields another equation (1'') which is true in case (1') is true
[ (1') being true is sufficient for (1'') being true ... ] :

((2Y)^(1/2))^5 = ((x^10 - z^5)^(1/5))^5 (1'')
((2Y)^(1/2))^5 = x^10 - z^5 (1''')

With x and z being integers, the right side of the
equation (1''') yields an integer.

(I'm not a mathematician, so the following thought might
need verification:) I think for the left side of the equation (1''')
to yield an integer, Y must be an even multiple of a square.
Otherwise the term (2Y)^(1/2) and thus also the term
((2Y)^(1/2))^5 which forms the left side of the equation will
not yield an element of the set of rational numbers and thus
also not an element of that subset of rational numbers which
is called "integers".

In other words: If a solution to (1) and thus also to (1''')
in integers existed, then there existed an integer B
with Y = 2B^2 :

((2*2B^2)^(1/2))^5 = x^10 - z^5 (1'''')

(2B)^5 = x^10 - z^5 (1''''')

(2B)^5 + z^5 = x^10 (1'''''')

(2B)^5 + z^5 = (x^2)^5 (1''''''')

Summa summarum: If a solution y, x, z in integers
each > 5 to (1) existed, then that solution could be
used for constructing a non-trivial integer-solution
[y = 2Y ; Y = 2B^2 -> y=4B^2 ->2B = sqrt(y)]
u=2B=sqrt(y) ; v=z; w=x^2
to u^n + v^n = w^n ; n=5 > 2
which - according to Fermat/Wiles - is not possible.

Thus I assume that there is no solution to (1)
where y, x, z are co prime integers each > 5, 2|y.

Sincerely

Ulrich

Date Subject Author
3/2/13 Deep Deb
3/2/13 quasi
3/2/13 Deep Deb
3/2/13 quasi
3/2/13 Deep Deb
3/14/13 Ulrich D i e z