Selection and Training Session

Question

Given nonzero real numbers a, b, c, with a + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3. (*) a) Find (1a + 1b + 1c) (a + b + c - 2). b) Do there exist pairwise different nonzero a, b, c satisfying (*)? (D. Bazylev)

Accepted Answer

a) (Solution of M.Mankevich, E.Dovgialo.) Let

σ_{1} = a + b + c

,

σ_{2} = ab + b c + c a

,

σ_{3} = ab c

. From the given equalities we have the system

σ_{1} = σ_{1}^{2} - 2 σ_{2} (1), σ_{1}^{2} = σ_{1}^{3} - 3 σ_{1} σ_{2} + 3 σ_{3} . (2)

From (1) we have

σ_{1}^{2} = σ_{1}^{3} - 2 σ_{1} σ_{2}

. Then from (2) it follows that

- 2 σ_{1} σ_{2} = - σ_{1} σ_{2} + 3 σ_{3} . (3)

Finally, the expression we search is

\frac{σ _{2}}{σ _{3}} (σ_{1} - 2) = \frac{σ _{1} σ _{2} - 2 σ _{2}}{σ _{3}} = [in view of (3)] = \frac{3 σ _{3}}{σ _{3}} = 3.

b) It is easy to verify, that

a = \frac{1}{2}

,

b = \frac{2 + 6}{4}

,

c = \frac{2 - 6}{4}

satisfy the given system.