396 Algebra
≤
1
27
(x+α+β+γ)^3 =1
27
(x−a)^3 ,so thatP(x)≥− 271 (x−a)^3. Equality holds exactly whenα−x=β−x=γ−xin
the first inequality andα+β+γ− 3 x=x+α+β+γin the second, that is, when
x=0 andα=β=γ.
Ifβ≤x≤γ, then using again the AM–GM inequality, we obtain
−P(x)=(x−α)(x−β)(γ−x)≤1
27
(x+γ−α−β)^3≤1
27
(x+α+β+γ)^3 =1
27
(x−a)^3 ,so that againP(x)≥− 271 (x−a)^3. Equality holds exactly when there is equality in both
inequalities, that is, whenα=β=0 andγ= 2 x.
Finally, whenα<x<βorx>γ, then
P(x) > 0 ≥−1
27
(x−a)^3.Thus the desired constant isλ=− 271 , and the equality occurs whenα=β=γand
x=0, or whenα=β=0,γis any nonnegative real, andx=γ 2.
(Chinese Mathematical Olympiad, 1999)
165.The key idea is to viewan+^1 −(a+ 1 )n−2001 as a polynomial ina. Its free term
is 2002, so any integer zero divides this number.
From here the argument shifts to number theory and becomes standard. First, note
that 2002= 2 × 7 × 11 ×13. Since 2001 is divisible by 3, we must havea≡ 1 (mod 3);
otherwise, one ofan+^1 and(a+ 1 )nwould be a multiple of 3 and the other not, and their
difference would not be divisible by 3. We deduce thata ≥7. Moreover,an+^1 ≡ 1
(mod 3), so we must have(a+ 1 )n≡ 1 (mod 3), which forcesnto be even, and in
particular at least 2.
Ifais even, thenan+^1 −(a+ 1 )n ≡−(a+ 1 )n (mod 4). Becausenis even,
−(a+ 1 )n≡− 1 (mod 4). But on the right-hand side, 2001≡ 1 (mod 4), and the
equality is impossible. Therefore,amust odd, so it divides 1001 = 7 × 11 ×13.
Moreover,an+^1 −(a+ 1 )n≡a(mod 4),soa≡ 1 (mod 4).
Of the divisors of 7× 11 ×13, those congruent to 1 modulo 3 are precisely those not
divisible by 11 (since 7 and 13 are both congruent to 1 modulo 3). Thusadivides 7×13.
Nowa≡ 1 (mod 4)is possible only ifadivides 13.
We cannot havea=1, since 1− 2 n =2001 for anyn. Hence the only possibility
isa=13. One easily checks thata=13,n=2 is a solution; all that remains to check
is that no othernworks. In fact, ifn>2, then 13n+^1 ≡ 2001 ≡ 1 (mod 8). But