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

>appreciated.

Many of your recent question are almost instantly resolved

as consequences of the following elementary lemma.

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.

proof;

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.

quasi