Date: Mar 28, 2013 3:51 PM
Author: quasi
Subject: Re: -----   ----- impossible equation

Deep wrote:

>Consider the following equation for the given conditions.
>a^(1/5).(2^m.T)^(1/5) = R^(1/2) (1)
>Conditions: a, m,T,R are integers each >1 such that a, T odd
>and R is non-square; a, T coprime.
>Conjecture: (1) has no solution.
>Any comment upon the correctness of the conjecture will be

Many of your recent question are almost instantly resolved
as consequences of the following elementary lemma.


If x,y,a,b are positive integers with gcd(a,b) = 1 such that

x^a = y^b

then x is a perfect b'th power and y is a perfect a'th power.


If x = 1, x^a = y^b implies y must also be equal to 1, so
in this case, the conclusion immediately follows.

Next suppose x > 1.

Let p be a prime factor of x and let p^j be the highest power
of p which divides x and let p^k be the highest power of p
which divides y. By the law of unique factorization,
p^(aj) = p^(bk), hence aj = bk. Since b|aj and gcd(a,b) = 1,
it follows that b|j. Thus, for every prime p factor if x, the
exponent of p in the prime factorization of x is a multiple
of b. It follows that x is a perfect b'th power, hence we can
write x = t^b for some positive integer t. Then the equation
x^a = y^b => t^(ab) = y^b => y = t^a, so x is a perfect b'th
power and y is a perfect a'th power, as was to be shown.

So now suppose

a^(1/5)*((2^m)*T)^(1/5) = R^(1/2)

where a,t,R are positive integers and m is a nonnegative
integer. Raising both sides to the 10'th power yields

(a*((2^m)*T))^2 = R^5

Letting x = a*((2^m)*T), we get x^2 = R^5, so the lemma implies
that R is a perfect square, contrary to your stated hypothesis.

Thus, there are no solutions.

Comment: Apparently, you don't follow the underlying logic in
the answers provided to you, since your question above is
essentially a repeat of recent questions you've asked.