(x, y) = ( 1, 1), (16, 2) and (27, 3) constitutes all possible solutions to the equation under reference.
METHOD 1:
Notice first that if we have am = bn, then we must have a = ce, b = cf, for some c, where m=fd, n=ed and d is the greatest common divisor of m and n. [Proof: express a and b as products of primes in the usual way.]
In this case let d be the greatest common divisor of a and b^2, and put a = de, b^2 = df. Then for some c, a = ce, b = cf. Hence :
f * (c^e) = e *(c2f). We cannot have e = 2f, for then the c's cancel to give e = f. Contradiction.
Suppose 2f > e, then f = e*c^(2f-e).
Hence e = 1 and f = c^(2f-1).
If c = 1, then f = 1 and we have the solution a = b = 1.
If c > = 2, then c^(2f-1) > = 2f > f, so there are no solutions.
Finally, suppose 2f < e. Then e = f.ce-2f.
Hence f = 1 and e = ce-2. ce-2 > = 2e-2 > = e for e> = 5, so we must have e = 3 or 4 (e > 2f = 2).
e = 3 gives the solution a = 27, b = 3. e = 4 gives the solution a = 16, b = 2.
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METHOD 2:
Let c be the greatest common divisor of a and b. Then, a = s*c and b = t*c, where c and t are relatively positive prime integers. Accordingly, the given equation reduces to
:
(s*c)^(c*(t^2)) = (c*t)^(s)
------------------------------------------(1)
We compare the coefficients in (i) in terms of the following cases.
CASE-I : If c *(t^2) = s; then (i) implies that s=t. Since s and t are relatively prime, it follows that s=t=1. This gives c=1. Hence (x, y) =(1,1) constitutes a solution to the problem.
CASE-II: If c*(t^2)> s, we will rewrite (i) in the form:
c^((c*(t^2) s ) * s^((c*(t^2) = t^s
-------------------- (2).
Accordingly, it follows that s^c*(t^2) divides t^s. Since t and s are relatively prime, we must have s =1. Consequently, we obtain from (2):
c^(c*(t^2) 1) = t
--------------------------------------(3)
If c=1, then from (3), we obtain t=1, so that the inequality corresponding to this case is violated.
If c > = 2 , then:
c^((c*(t^2) - 1)) > = 2^(2*(t^2) - 1)) > t , for any positive integer t. This violates (3) and, accordingly, this is a contradiction.
CASE-III: If c*(t^2)< s we can rewrite (1) as :
s^(c*(t^2)) = (c^(s (c*(t^)) * (t^s)
--------------------(4)
and note that t^s divides s^(c*(t^2)). Since, s and t are relatively prime, it follows that t=1, and accordingly, (4) reduces to:
s^c = c^(s c)
--------------------------------------(5)
Now, c is clearly less than s, and accordingly, from (5), we obtain:
c < s c; giving s > 2*c, which is in conformity with the tenets constituting relationship (5).
Consequently, c divides s, so that s = k*c, for some positive integer k. Hence, substituting s = k*c in (5), we obtain: k = c^(k - 2)
--------------------------------------(6)
Clearly, the RHS of the abovementioned relationship is inclusive of positive integral values only when k > = 2.
Now, for k =2, we obtain 2 = 1, which is a contradiction.
But, we observe that , c^(k - 2) > = 2^(k - 2)> k, for all k greater than or equal to 5.
Accordingly, k = 3, 4.
For k =3, we obtain c=3, giving s=k*c =9 or, a =27; b = 3
For k = 4, we obtain c=2, giving s = 8; so that a = 16; b = 2.
Hence, (a, b) = ( 1, 1), (16, 2) and (27, 3) constitutes all possible solutions to the equation under reference.
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