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.

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