There exist a real number p
such that the following system of equations
1 + 2 x 2 1 + 1 + 2 y 2 1 = 1 + 2 x y 2
x x ( 1 − 2 x ) + y ( 1 − 2 y ) = p
has only one real value x as a root, which can be expressed as
x = c a + b
where a , b , c are all non-zero integers and b is square-free. Find a + b + c
Details and Assumptions :
See Victor Loh's Irrational Equations
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If we look from the first equation we can say that x=y is a solution, which make the second equation after we change y by x : -2x^4-3x^3+x-p^2=0 the problem here is what is the value of p can make the equation have 1 solution so we need simply to find the max of this equation -2x^4-3x^3+x by using the derivative of that fonction so we'll have -8x^3 -9x^2+1=0 that's how you can find p
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For
p = 6 4 1 − 1 1 7 + 1 6 5 3 3 = 0 . 4 5 0 3 8 3 . . .
the only real root is
x = 1 6 − 1 + 3 3 .
Thus, the answer is 4 8 .