Advanced book on Mathematics Olympiad

Algebra 363

We further change this intox =^3

√

x^3 −x. Raising both sides to the third power, we
obtainx^3 =x^3 −x. We conclude that the equation has the unique solutionx=0.

Second solution: The functionf:R→R,f(x)=^3

√

x− 1 +^3

√

x+^3

√

x+1 is strictly
increasing, so the equationf(x)=0 has at most one solution. Sincex=0 satisfies this
equation, it is the unique solution.

91.The key observation is that the left-hand side of the equation can be factored as

(x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx)=p.

Sincex+y+z>1 andpis prime, we must havex+y+z=pandx^2 +y^2 +z^2 −xy−
yz−zx=1. The second equality can be written as(x−y)^2 +(y−z)^2 +(z−x)^2 =2.
Without loss of generality, we may assume thatx≥y≥z.Ifx>y>z, thenx−y≥1,
y−z≥1, andx−z≥2, which would imply that(x−y)^2 +(y−z)^2 +(z−x)^2 ≥ 6 >2.
Therefore, eitherx=y=z+1orx− 1 =y=z. According to whether the prime
pis of the form 3k+1or3k+2, the solutions are(p− 31 ,p− 31 ,p+ 32 )and the corresponding
permutations, or(p− 32 ,p+ 31 ,p+ 31 )and the corresponding permutations.
(T. Andreescu, D. Andrica,An Introduction to Diophantine Equations, GIL 2002)

92.The inequality to be proved is equivalent to

a^3 +b^3 +c^3 − 3 abc≥ 9 k.

The left-hand side can be factored, and the inequality becomes

(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)≥ 9 k.

Without loss of generality, we may assume thata≥b≥c. It follows thata−b≥1,
b−c≥1,a−c≥2; hence(a−b)^2 +(b−c)^2 +(c−a)^2 ≥ 1 + 1 + 4 =6. Dividing
by 2, we obtain

a^2 +b^2 +c^2 −ab−bc−ca≥ 3.

The solution will be complete if we show thata+b+c≥ 3 k. The computation

(a+b+c)^2 =a^2 +b^2 +c^2 −ab−bc−ca+ 3 (ab+bc+ca)

≥ 3 + 3 ( 3 k^2 − 1 )= 9 k^2

completes the proof.
(T. Andreescu)

93.This is a difficult exercise in completing squares. We have

mnp= 1 +

x^2

z^2

+

z^2

y^2

+

x^2

y^2

+

y^2

x^2

+

y^2

z^2

+

z^2

x^2

+ 1