404 Algebra
Returning to the problem, we see that ify 0 andy 1 are zeros ofP′(x)−kP (x)with
y 0 <y 1 , then they are separated by a zero ofPand satisfyf(y 0 )=f(y 1 )=k. From
the lemma it follows that we cannot havey 1 ≤y 0 +δ(P(x)),soy 1 −y 0 >d, and we
are done.
(American Mathematical Monthly, published in a note by P. Walker, solution by
R. Gelca)
182.The number 101 is prime, yet we cannot apply Eisenstein’s criterion because of the
- The trick is to observe that the irreducibility ofP(x)is equivalent to the irreducibility
ofP(x− 1 ). Because the binomial coefficients
( 101
k)
,1≤k≤100, are all divisible by
101, the polynomialP(x− 1 )has all coefficients but the first divisible by 101, while
the last coefficient is(− 1 )^101 + 101 (− 1 )^101 + 102 =202, which is divisible by 101 but
not by 101^2. Eisenstein’s criterion proves thatP(x− 1 )is irreducible; henceP(x)is
irreducible as well.
183.Note thatP(x)=(xp− 1 )/(x− 1 ).IfP(x)were reducible, then so would be
P(x+ 1 ). But
P(x+ 1 )=
(x+ 1 )p− 1
x=xp−^1 +(
p
1)
xp−^1 +···+(
p
p− 1)
.
The coefficient
(p
k)
is divisible bypfor all 1≤k≤p−1, and( p
p− 1)
=pis not divisible
byp^2 ; thus Eisenstein’s criterion applies to show thatP(x+ 1 )is irreducible. It follows
thatP(x)itself is irreducible, and the problem is solved.
184.Same idea as in the previous problem. We look at the polynomial
P(x+ 1 )=(x+ 1 )^2n
+ 1=x^2
n
+(
2 n
1)
x^2
n− 1
+(
2 n
2)
x^2n− (^1) − 2
+···+
(
2 n
2 n− 1)
x+ 2.For 1≤k≤ 2 n, the binomial coefficient
( 2 n
k)
is divisible by 2. This follows from the
equality
(
2 n
k
)
=
2 n
k(
2 n− 1
k− 1)
,
since the binomial coefficient on the right is an integer, and 2 appears to a larger power
in the numerator than in the denominator. The application of Eisenstein’s irreducibility
criterion is now straightforward.
185.Arguing by contradiction, assume thatP(x)can be factored, and letP(x)=
Q(x)R(x). BecauseP(ai)=− 1 ,i= 1 , 2 ,...,n, andQ(ai)andR(ai)are integers,
eitherQ(ai)=1 andR(ai)=−1, orQ(ai)=−1 andR(ai)=1. In both situations