360 Algebra
F(x)=(P 1 (x)^2 +Q 1 (x)^2 )(P 2 (x)^2 +Q 2 (x)^2 )···(Pn(x)^2 +Qn(x)^2 ),where the factorR(x)^2 is incorporated inP 1 (x)^2 andQ 1 (x)^2. Using the Lagrange identity
(a^2 +b^2 )(c^2 +d^2 )=(ac+bd)^2 +(ad−bc)^2 ,we can transform this product in several steps intoP(x)^2 +Q(x)^2 , whereP(x)andQ(x)
are polynomials.
Second solution: Likewise, with the first solution write the polynomial as
F(x)=R(x)^2 (x^2 +a 1 x+b 1 )(x^2 +a 2 x+b 2 )···(x^2 +anx+bn).Factor the quadratics as(x+αk+iβk)(x+αk−iβk). Group the factors with+iβkinto a
polynomialP(x)+iQ(x)and the factors with−iβkinto the polynomialP(x)−iQ(x).
Then
F(x)=(R(x)P (x))^2 +(R(x)Q(x))^2 ,which proves the conclusion.
Remark.D. Hilbert discovered that not every positive two-variable polynomial can be
written as a sum of squares of polynomials. The appropriate generalization to the case
of rational functions makes the object of his 16th problem. While Hilbert’s proof is
nonconstructive, the first examples of such polynomials were discovered surprisingly late,
and were quite complicated. Here is a simple example found by T. Motzkin:f (x, y)=
1 +x^2 y^2 (x^2 +y^2 − 3 ).
84.Simply substitutex= 55
n
in the factorization
x^5 +x+ 1 =(x^2 +x+ 1 )(x^3 −x^2 + 1 )to obtain a factorization of the number from the statement. It is not hard to prove that
both factors are greater than 1.
(T. Andreescu, published in T. Andreescu, D. Andrica, 360Problems for Mathematical
Contests, GIL, 2003)
85.Let
N= 5 n−^1 −(
n
1)
5 n−^2 +(
n
2)
5 n−^3 −···+(
n
n− 1)
.
Then 5N− 1 =( 5 − 1 )n. Hence
N=
4 n+ 1
5=
4 ( 2 k)^4 + 1
5=
( 22 k+^1 + 2 k+^1 + 1 )( 22 k+^1 − 2 k+^1 + 1 )
5