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(x^2+y^2)^2 + 2tx(x^2 + y^2) = t^2y^2

Source: Turkey TST 2003 - P4

April 6, 2013
functionalgebra proposedalgebranumber theory

Problem Statement

Find the least
a. positive real number
b. positive integer
tt such that the equation (x2+y2)2+2tx(x2+y2)=t2y2(x^2+y^2)^2 + 2tx(x^2 + y^2) = t^2y^2 has a solution where x,yx,y are positive integers.
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