# Does there exist a triple of $distinct$ numbers $a,b,c$ such that $(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$?

Does there exist a triple of distinct numbers $a,b,c$ such that $$(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$$ ?

SOURCE : Inequalities (PDF) (Page Number 4 ; Question Number 220.1)

I tried expanding the brackets and I ended up with this messy equation :

$$-5 a^4 b + 5 a^4 c + 10 a^3 b^2 - 10 a^3 c^2 - 10 a^2 b^3 + 10 a^2 c^3 + 5 a b^4 - 5 a c^4 - 5 b^4 c + 10 b^3 c^2 - 10 b^2 c^3 + 5 b c^4 = 0$$

There is no hope of setting $a=b$ or $a=c$ as the question specifically asks for distinct numbers. So, at last I started collecting, grouping, factoring and manipulating the terms around but could find nothing. Wolfram|Alpha gives a solution as :

$$c=\dfrac{1}{2}\big(\pm\sqrt{3}\sqrt{-(a-b)^2} + a+b\big)$$

How can this solution be found?

Another thing I notice about the solution is that it contains a negative term inside the square root, so does that mean that the solution involves complex numbers and that there is no solution for $\big(a,b,c\big)\in \mathbb {R}$ ?

I am very confused about how to continue. Can anyone provide a solution/hint on how to 'properly' solve this problem ?

• If complex numbers are included, then this is possible, otherwise $x^2+xy+y^2>0$. – Wolfram Feb 12 '17 at 14:37

Assume that $a,b,c$ are distinct.

Let $a-b=x \neq 0,\; b-c=y \neq 0$. Note that $a-c=x+y \neq 0$

Note that the equation becomes $$x^5+y^5=(x+y)^5$$ So $$(x+y)^5-x^5-y^5=5xy(x^3+2x^2y+2xy^2+y^3)=0$$

Note that this becomes $$xy(x+y)(x^2+y^2+xy)=0 \iff x^2+xy+y^2=0$$

Using the quadratic formula, we can find $x,y$. Note that there are only non-real solutions.

• Thanks for this amazing answer !!! Truly ingenious ! :) +1 – user399078 Feb 12 '17 at 14:49