520 Real Analysis
Thus the lack of solutions will follow if we show thatFis strictly increasing. Recall that
e−t> 1 −tfort =0, hence 2−z> 1 −zlog 2 forz =0. Hence
F′(z)=
2 z(zlog 2− 1 + 2 −z)
z^2> 0
forz =0 and henceFis strictly increasing.
(T. Andreescu, second solution by R. Stong)
422.Clearly,αis nonnegative. Define f (x)=f(x+ 1 )−f(x), and (k)f(x)=
( (k−^1 )f(x)),k ≥ 2. By the mean value theorem, there existsθ 1 ∈ ( 0 , 1 )such
f(x+ 1 )−f(x)=f′(x+θ 1 ), and inductively for everyk, there existsθk∈( 0 ,k)such
that (k)f(x)=f(k)(x). Applying this tof(x)=xαandx=n, we conclude that for
everykthere existsθk∈( 0 ,k)such thatf(k)(n+θk)is an integer. Choosek=α+1.
Then
(k)f(n+θ)=α(α− 1 )···(α+ 1 −k)
(n+θk)k−α.
This number is an integer by hypothesis. It is not hard to see that it is also positive and
less than 1. The only possibility is that it is equal to 0, which means thatα=k−1, and
the conclusion follows.
(W.L. Putnam Mathematical Competition)
423.The equation isa^3 +b^3 +c^3 = 3 abc, witha= 2 x,b=− 3 x−^1 , andc=−1. Using
the factorization
a^3 +b^3 +c^3 − 3 abc=1
2
(a+b+c)[
(a−b)^2 +(b−c)^2 +(c−a)^2]
we find thata+b+c=0 (the other factor cannot be zero since, for example, 2xcannot
equal−1). This yields the simpler equation
2 x= 3 x−^1 + 1.Rewrite this as
3 x−^1 − 2 x−^1 = 2 x−^1 − 1.We immediately notice the solutionsx=1 andx=2. Assume that another solution
exists, and consider the functionf(t)=tx−^1. Becausef( 3 )−f( 2 )=f( 2 )−f( 1 ),by
the mean value theorem there existt 1 ∈( 2 , 3 )andt 2 ∈( 1 , 2 )such thatf′(t 1 )=f′(t 2 ).
This gives rise to the impossible equality(x− 1 )t 1 x−^2 =(x− 1 )t 2 x−^2. We conclude that
there are only two solutions:x=1 andx=2.
(Mathematical Reflections, proposed by T. Andreescu)