688 Number Theory
for some integersa, b, c. Identifying coefficients, we must have
a+b=− 4 ,
c+ab+ 1 = 6 ,
b+ac+ 1 =− 4 ,
a+c= 1.From the first and last equations, we obtainb−c=−5, and from the second and the
third,(b−c)(a− 1 )=10. It follows thata− 1 =−2; hencea=−1,b=− 4 + 1 =−3,
c= 1 + 1 =2. Therefore,
n(n− 1 )^4 + 1 =(n^2 −n+ 1 )(n^3 − 3 n^2 + 2 n+ 1 ),a product of integers greater than 1.
(T. Andreescu)
732.Settingn=0 in (i) gives
f( 1 )^2 =f( 0 )^2 + 6 f( 0 )+ 1 =(f ( 0 )+ 3 )^2 − 8.Hence
(f ( 0 )+ 3 )^2 −f( 1 )^2 =(f ( 0 )+ 3 +f( 1 ))(f ( 0 )+ 3 −f( 1 ))= 4 × 2.The only possibility isf( 0 )+ 3 +f( 1 )=4 andf( 0 )+ 3 −f( 1 )=2. It follows that
f( 0 )=0 andf( 1 )=1.
In general,
(f ( 2 n+ 1 )−f( 2 n))(f ( 2 n+ 1 )+f( 2 n))= 6 f (n)+ 1.We claim thatf( 2 n+ 1 )−f( 2 n)=1 andf( 2 n+ 1 )+f( 2 n)= 6 f(n)+1. To prove our
claim, letf( 2 n+ 1 )−f( 2 n)=d. Thenf( 2 n+ 1 )+f( 2 n)=d+ 2 f( 2 n). Multiplying,
we obtain
6 f (n)+ 1 =d(d+ 2 f( 2 n))≥d(d+ 2 f (n)),where the inequality follows from condition (ii). Moving everything to one side, we
obtain the inequality
d^2 +( 2 d− 6 )f (n)− 1 ≤ 0 ,which can hold only ifd≤3. The casesd=2 andd=3 cannot hold, becauseddivides
6 f (n)+1. Henced=1, and the claim is proved. From it we deduce thatfis computed
recursively by the rule