quasi
Posts:
10,970
Registered:
7/15/05


Re:   impossible equation
Posted:
Mar 28, 2013 3:51 PM


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 nonsquare; 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 baj and gcd(a,b) = 1, it follows that bj. 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

