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 Find Five Fractions (Posted on 2016-03-27)
There are already two puzzles on this site asking to find three real roots to an equation of the form:
(x^3+k*x)^(1/5) = (x^5-k*x)^(1/3)

The two puzzles are k=2 here and k=20 here.

I challenge you to find a rational value of k such that there are four nonzero rational roots to the equation above.

 No Solution Yet Submitted by Brian Smith Rating: 4.0000 (2 votes)

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 Heuristic approach - spoiler | Comment 1 of 2
Using a heuristic/mystic approach….

Let        (x3 +kx)1/5 = (x5 – kx)1/3 = t

then      x3 + kx = t5       and       x5 – kx = t3

Now let t = x. Who knows, it might produce a family of
possibilities and, if so, they will satisfy both equations
since this substitution makes the equations identical.

viz.       x5 – x3 – kx = 0

We’re only interested in non-zero solutions so divide by x.

x4 – x2 – k = 0                                       (1)

Now we need this equation to have four rational roots.
If one of its roots is r, then k = r4 - r2, and the equation becomes

x4 – x2 – r4 + r2 = 0     which factorises to:

(x – r)(x + r)(x2 + r2 – 1) = 0

giving the four roots:     +/- r  and  +/- sqrt(1 – r2)

So providing r and sqrt(1 – r2) are rational, we have four solutions.

Choose any primitive Pythagorean triple, {a, b, c} with a2 + b2 = c2,

then let  r = a/c and the four roots are:    a/c, -a/c, b/c, -b/c.

Then from (1)       k = - product of roots  =  - a2b2/c4

For example:     Using {3, 4, 5}, k = - 144/625.

 Posted by Harry on 2016-03-27 20:00:53

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