Advanced book on Mathematics Olympiad

Number Theory 675

−an≤−b^2 n≤−an− 1 ,

an≤b^2 n+ 1 ≤an+ 1.

Adding these two inequalities, we obtain

0 ≤bn^2 + 1 −bn^2 ≤bn+bn− 1 ,

or

0 ≤(bn+ 1 −bn)(bn+ 1 +bn)≤bn+bn− 1.

Therefore,bn+ 1 ≥bnforn= 2 , 3 ,4. If forn=3orn=4 this inequality were strict,

then for that specificnwe would have

0 <b^2 n+ 1 −b^2 n≤bn+bn− 1 <bn+ 1 +bn,

with the impossible consequence 0<bn+ 1 −bn <1. It follows thatb 3 =b 4 =b 5.

Combining this with the inequality from the statement, namely with

b^23 ≤a 3 ≤b^24 ≤a 4 ≤b^25 ,

we find thata 3 =a 4. But thenb 3 =a 4 −a 3 =0, which would implya 2 ≤b^23 =0, a

contradiction. We conclude that the sequence can have at mostfiveterms. This limit is

sharp, sincea 1 =1,a 2 =3,a 3 =4,a 4 =6,a 5 =8 satisfies the condition from the

statement.

(Romanian Team Selection Test for the International Mathematical Olympiad, 1985,

proposed by L. Panaitopol)

705.Settingx=y=z=0 we find thatf( 0 )= 3 (f ( 0 ))^3. This cubic equation has

the unique integer solutionf( 0 )=0. Next, withy=−xandz=0 we havef( 0 )=

(f (x))^3 +(f (−x))^3 +(f ( 0 ))^3 , which yieldsf(−x)=−f(x)for all integersx; hencef

is an odd function. Now setx=1,y=z=0 to obtainf( 1 )=(f ( 1 ))^3 + 2 (f ( 0 ))^3 ; hence

f( 1 )=f( 1 )^3. Therefore,f( 1 )∈{− 1 , 0 , 1 }. Continuing withx=y=1 andz=0 and

x=y=z=1 we find thatf( 2 )= 2 (f ( 1 ))^3 = 2 f( 1 )andf( 3 )= 3 (f ( 1 ))^3 = 3 f( 1 ).

We conjecture thatf(x)=xf ( 1 )for all integersx. We will do this by strong induction

on the absolute value ofx, and for that we need the following lemma.

Lemma.Ifxis an integer whose absolute value is greater than 3 , thenx^3 can be written

as the sum of five cubes whose absolute values are less thanx.

Proof.We have

43 = 33 + 33 + 23 + 13 + 13 , 53 = 43 + 43 +(− 1 )^3 +(− 1 )^3 +(− 1 )^3 ,

63 = 53 + 43 + 33 + 03 + 03 , 73 = 63 + 53 + 13 + 13 + 03 ,