Copyright © University of Cambridge. All rights reserved.

'Weekly Challenge 39: Symmetrically So' printed from http://nrich.maths.org/

Show menu


In this case make the substitution $y=x+4$, which transforms the equation into
$$
(y-1)^4+(y+1)^4=20.
$$
This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.
 
If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:
$$
(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.
$$
So, the odd factors cancel to give
$$
2y^4+12y^2-18=0
$$
This then becomes the simple equation
$$
y^4+6y^2-9=0\,,
$$
which is a quadratic equation in $y^2$ with solutions
$$
y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).
$$
One of these solutions is positive; taking the square root gives two real values for $y$ as
$$
y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}
$$
Therefore two real solutions to the equation are 
$$
x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}
$$