Methods of Proof 32533 =f( 2 )^3 =f( 23 )<f( 32 )=f( 3 )^2 =k^2 ,hencek>5. Similarly, using 3^3 < 25 , we obtain
k^3 =f( 3 )^3 =f( 33 )<f( 25 )=f( 2 )^5 = 35 = 243 < 343 = 73.This implies thatk<7, and consequentlykcan be equal only to 6. Thus we should
havef( 2 )=3 andf( 3 )=6. The monotonicity offimplies that 2u< 3 vif and only
if 3u< 6 v,u, vbeing positive integers. Taking logarithms this means thatvu>log 2 3if
and only ifvu >log 3 6. Since rationals are dense, it follows that log 23 =log 3 6. This
can be written as log 23 =log^123 +1, and so log 2 3 is the positive solution of the quadratic
equationx^2 −x− 1 =0, which is the golden ratio^1 +
√
5
2. The equality21 + 2 √ 5
= 3translates to 2^1 +
√(^5) =9. But this would imply
65536 = 25 ×^3.^2 < 25 (^1 +
√
5 )= 95 = 59049.
We have reached a contradiction, which proves that the functionfcannot exist.
(B.J. Venkatachala,Functional Equations: A Problem Solving Approach, Prism
Books PVT Ltd., Bangalore, 2002)
8.The constant functionf(x)=k, wherekis a positive integer, is the only possible
solution. That any such function satisfies the given condition is easy to check.
Now suppose there exists a nonconstant solutionf. There must exist two positive
integersaandbsuch thatf(a)<f(b). This implies that(a+b)f(a) < af(b)+bf ( a ) <
(a+b)f (b), which by the given condition is equivalent to(a+b)f(a)<(a+b)f (a^2 +
b^2 )<(a+b)f (b). We can divide bya+b>0 to find thatf(a)<f(a^2 +b^2 )<f(b).
Thus between any two different values offwe can insert another. But this cannot go on
forever, sinceftakes only integer values. The contradiction shows that such a function
cannot exist. Thus constant functions are the only solutions.
(Canadian Mathematical Olympiad, 2002)
9.Assume thatA,B, andasatisfyA∪B=[ 0 , 1 ],A∩B=∅,B=A+a. We can assume
thatais positive; otherwise, we can exchangeAandB. Then( 1 −a, 1 ]⊂B; hence
( 1 − 2 a, 1 −a]⊂A. An inductive argument shows that for any positive integern, the
interval( 1 −( 2 n+ 1 )a, 1 − 2 na]is inB, while the interval( 1 −( 2 n+ 2 )a, 1 −( 2 n+ 1 )a]
isinA. However, at some point this sequence of intervals leaves[ 0 , 1 ]. The interval of
the form( 1 −na, 1 −(n− 1 )a]that contains 0 must be contained entirely in eitherAor
B, which is impossible since this inteval exits[ 0 , 1 ]. The contradiction shows that the
assumption is wrong, and hence the partition does not exist.
(Austrian–Polish Mathematics Competition, 1982)