Algebra 405(Q+R)(ai)=0,i = 1 , 2 ,...,n. Since theai’s are all distinct and the degree of
Q(x)+R(x)is at mostn−1, it follows thatQ(x)+R(x)≡0. HenceR(x)=−Q(x),
andP(x)=−Q^2 (x). But this contradicts the fact that the coefficient of the term of
maximal degree inP(x)is 1. The contradiction proves thatP(x)is irreducible.
(I. Schur)
186.Assume that the polynomialP(x)is reducible, and write it as a productQ(x)R(x)of
monic polynomials with integer coefficients of degreei, respectively, 2n−i. BothQ(x)
andR(x)are positive for any real numberx(being monic and with no real zeros), and from
Q(ak)R(ak)=1,k= 1 , 2 ,...,n, we find thatQ(ak)=R(ak)=1,k= 1 , 2 ,...,n.
If, say,i<n, then the equationQ(x)=1 hasnsolutions, which, taking into account
the fact thatQ(x)has degree less thann, means thatQ(x)is identically equal to 1. This
contradicts our original assumption. Also, ifi=n, the polynomialQ(x)−R(x)hasn
zeros, and has degree less thann, so it is identically equal to 0. Therefore,Q(x)=R(x),
which means that
(x−a 1 )^2 (x−a 2 )^2 ···(x−an)^2 + 1 =Q(x)^2.Substituting integer numbers forx, we obtain infinitely many equalities of the form
p^2 + 1 =q^2 , withpandqintegers. But this equality can hold only ifp=0 andq=1,
and we reach another contradiction. Therefore, the polynomial is irreducible.
(I. Schur)
187.LetP(x)=anxn+an− 1 xn−^1 +···+a 0 , and assume to the contrary thatP(x)=
Q(x)R(x), whereQ(x)andR(x)are polynomials with integer coefficients of degree
at least 1 (the degree zero is ruled out because any factor that divides all coefficients of
P(x)divides the original prime).
Because the coefficients ofP(x)are nonnegative integers between 0 and 9, and the
leading coefficient is positive, it follows that the zeros ofP(x)are in the union of the left
half-plane Imz≤0 and the disk|z|<4. Otherwise, if Imz>0 and|z|≥4, then
1 ≤an≤Re(an+an− 1 z−^1 )=Re(−a 2 z−^2 −···−anz−n)<9 |z|−^2
1 −|z|−^1≤
3
4
,
a contradiction.
On the other hand, by hypothesisP( 10 )is prime; hence eitherQ( 10 )orR( 10 )is
1 (or−1 but then just multiply both polynomials by−1). AssumeQ( 10 )=1, and let
Q(x)=c(x−x 1 )(x−x 2 )···(x−xk). Thenxi,i= 1 , 2 ,...,k, are also zeros ofP(x),
and we have seen that these lie either in the left half-plane or in the disk of radius 4
centered at the origin. It follows that
1 =Q( 10 )=|Q( 10 )|=|c|·| 10 −x 1 |·| 10 −x 2 |···| 10 −xk|≥|c|· 6 k,