Advanced book on Mathematics Olympiad

434 Algebra

1 + 2 +···+(n− 1 )+ 1 + 2 +···+n=

n(n− 1 )

2

+

n(n+ 1 )

2

=n^2.

The problem is solved.

Remark.It is interesting to note that the same method allows the computation of the

inverse as(bk,m)km, giving

bk,m=

(− 1 )k+m(n+k− 1 )!(n+m− 1 )!

(k+m− 1 )[(k− 1 )!(m− 1 )!]^2 (n−m)!(n−k)!

.

241.First, note that the polynomials

(x

1

)

,

(x+ 1

3

)

,

(x+ 2

5

)

,...are odd and have degrees

1 , 3 , 5 ,...,and so they form a basis of the vector space of the odd polynomial func-

tions with real coefficients.

The scalarsc 1 ,c 2 ,...,cmare computed successively from

P( 1 )=c 1 ,

P( 2 )=c 1

(

2

1

)

+c 2 ,

P( 3 )=c 1

(

3

1

)

+c 2

(

4

3

)

+c 3.

The conclusion follows.

(G. Pólya, G. Szego, ̋Aufgaben und Lehrsätze aus der Analysis, Springer-Verlag, 1964)

242.Inspired by the previous problem we consider the integer-valued polynomials

(x

m

)

=

x(x− 1 )···(x−m+ 1 )/m!,m= 0 , 1 , 2 ,....They form a basis of the vector space of

polynomials with real coefficients. The system of equations

P(k)=b 0

(

x

n

)

+b 1

(

x

n− 1

)

+···+bn− 1

(

x

1

)

+bn,k= 0 , 1 ,...,n,

can be solved by Gaussian elimination, producing an integer solutionb 0 ,b 1 ,...,bn.
Yes, we do obtain an integer solution because the coefficient matrix is triangular and has
ones on the diagonal! Finally, when multiplying

(x

m

)

,m= 0 , 1 ,...,n,byn!, we obtain

polynomials with integer coefficients. We find thatn!P(x)has integer coefficients, as

desired.

(G. Pólya, G. Szego, ̋Aufgaben und Lehrsätze aus der Analysis, Springer-Verlag, 1964)

243.Forn=1 the rank is 1. Let us consider the casen≥2. Observe that the rank does

not change under row/column operations. Fori=n, n− 1 ,...,2, subtract the(i− 1 )st

row from theith. Then subtract the second row from all others. Explicitly, we obtain