674 Number Theory
701.Fromf( 1 )+ 2 f(f( 1 ))=8 we deduce thatf( 1 )is an even number between 1 and 6,
that is,f( 1 )= 2 ,4, or 6. Iff( 1 )=2 then 2+ 2 f( 2 )=8, sof( 2 )=3. Continuing with
3 + 2 f( 3 )=11, we obtainf( 3 )=4, and formulate the conjecture thatf (n)=n+ 1
for alln≥1. And indeed, in an inductive manner we see thatf (n)=n+1 implies
n+ 1 + 2 f(n+ 1 )= 3 n+5; hencef(n+ 1 )=n+2.
The casef( 1 )=4 gives 4+ 2 f( 4 )= 8,sof( 4 )=2. But then 2+ 2 f(f( 4 ))=17,
which cannot hold for reasons of parity. Also, iff( 1 )=6, then 6+ 2 f( 6 )=8, so
f( 6 )=1. This cannot happen, becausef( 6 )+ 2 f(f( 6 ))= 1 + 2 ·6, which does not
equal 3· 6 +5.
We conclude thatf (n)=n+1,n≥ 1, is the unique solution to the functional
equation.
702.Letg(x)=f(x)−x. The given equation becomesg(x)= 2 g(f (x)). Iterating,
we obtain thatg(x)= 2 nf(n)(x)for allx ∈Z, wheref(n)(x)meansf composedn
times with itself. It follows that for everyx∈Z,g(x)is divisible by all powers of 2,
sog(x)=0. Therefore, the only function satisfying the condition from the statement is
f(x)=xfor allx.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
L. Funar)
703.Assume such a function exists, and defineg:N→ 3 N+1,g(x)= 3 f(x)+1. Then
gis bijective and satisfiesg(xy)=g(x)g(y). This implies in particular thatg( 1 )=1.
We will need the following fact. Ifxis such thatg(x)=n, wheren=pq, andp, q
are prime numbers congruent to 2 modulo 3, thenxis prime. Indeed, ifx=yz,y, z≥2,
theng(x)=g(y)g(z). This implies thatncan be factored as the product of two numbers
in 3N+1, which is not true.
Now choose two distinct numberspandqthat are congruent to 2 modulo 3 (for
example, 2 and 5). Thenpq,p^2 , andq^2 are all in the image ofg. Letg(a)=p^2 ,
g(b)=q^2 , andg(c)=pq. We have
g(c^2 )=g(c)^2 =p^2 q^2 =g(a)g(b)=g(ab).It follows thatc^2 =ab, witha, b, andcdistinct prime numbers, and this is impossible.
Therefore, such a functionfdoes not exist.
(Balkan Mathematical Olympiad, 1991)
704.We will prove that a sequence of positive integers satisfying the double inequality
from the statement terminates immediately. Precisely, we show that ifa 1 ,a 2 ,...,aN
satisfy the relation from the statement forn= 1 , 2 ,...,N, thenN≤5.
Arguing by contradiction, let us assume that the sequence has a sixth terma 6. Set
bn=an+ 1 −an,n= 1 ,...,5. The relation from the statement impliesan≥an− 1 for
n≥2, and sobnis a nonnegative integer forn= 1 ,...,5. Forn= 2 , 3 ,4 we have