Advanced book on Mathematics Olympiad

64 2 Algebra

Solution.The key idea is to viewxnas a variable and think of the determinant as an
(n− 1 )st-degree polynomial inxn. The leading coefficient is itself a Vandermonde
determinant of ordern−1, while then−1 roots are obviouslyx 2 ,x 3 ,...,xn− 1. The
determinant is therefore equal to
∣
∣∣
∣∣
∣∣
∣
∣

11 ··· 1

x 1 x 2 ···xn− 1

..

.

..

.

... ..

.

xn 1 −^2 x 2 n−^2 ···xnn−− 12

∣

∣∣

∣∣

∣∣

∣

∣

(xn−x 1 )(xn−x 2 )···(xn−xn− 1 ).

Now we can induct on∏ nto prove that the Vandermonde determinant is equal to

i>j(xi−xj). This determinant is equal to zero if and only if two of thexi’s are
equal. 

We continue with a problem of D. Andrica.

Example.(a) Consider the real numbersaij,i= 1 , 2 ,...,n−2,j= 1 , 2 ,...,n,n≥3,
and the determinants

Ak=

∣∣

∣∣

∣∣

∣

∣∣

1 ··· 11 ··· 1

a 11 ··· a 1 ,k− 1 a 1 ,k+ 1 ··· a 1 n

..

.

... ..

.

..

.

... ..

.

an− 2 , 1 ···an− 2 ,k− 1 an− 2 ,k+ 1 ···an− 2 ,n

∣∣

∣∣

∣∣

∣

∣∣

.

Prove that

A 1 +A 3 +A 5 +··· =A 2 +A 4 +A 6 +···.

(b) Define

pk=

n−∏(k+ 1 )

i= 0

(xn−i−xk), qk=

k∏− 1

i= 1

(xk−xi),

wherexi,i= 1 , 2 ,...,n, are some distinct real numbers. Prove that

∑n

k= 1

(− 1 )k

pkqk

= 0.

(c) Prove that for any positive integern≥3 the following identity holds:

∑n

k= 1

(− 1 )kk^2

(n−k)!(n+k)!

= 0.