```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^bthen 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, soin 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 canwrite x = t^b for some positive integer t. Then the equationx^a = y^b => t^(ab) = y^b => y = t^a, so x is a perfect b'thpower 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 nonnegativeinteger. Raising both sides to the 10'th power yields   (a*((2^m)*T))^2 = R^5Letting x = a*((2^m)*T), we get x^2 = R^5, so the lemma impliesthat R is a perfect square, contrary to your stated hypothesis.Thus, there are no solutions.Comment: Apparently, you don't follow the underlying logic inthe answers provided to you, since your question above isessentially a repeat of recent questions you've asked.quasi
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