
Re:     validity of an equation
Posted:
Mar 2, 2013 1:25 PM


On Saturday, March 2, 2013 8:43:43 AM UTC5, Deep wrote: > Consider the following equation under the given conditions > > > > y^(1/2) = (x^10  z^ 5)^(1/5 ) (1) > > > > Conditions: y, x, z are co prime integers each > 5, 2y. > > > > Question: can (1) have any solution ? > > > > Any helpful comment will be appreciated
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Thank you very much for your clear reply. Now consider the situation where y is not a perfect square. Then from y^5 = w^2 implies y^(5/2) = w. Then w can not be an integer. This confirms your conclusion without FLT
Kindly comment.
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