408 Algebra
Now let us assume thatP(x)is a polynomial for whichM=^18. Thenb=−1,
d=^18. Writing the double inequality−^18 ≤P(x)≤^18 forx=1 and−1, we obtain
−^18 ≤^18 +a+c≤^18 and−^18 ≤^18 −a−c≤^18. So on the one hand,a+c≥0,
and on the other hand,a+c≤ 0. It follows thata =−c. But then forx = √^12 ,
0 ≤a( 2 √^12 −√^12 )≤^14 , and forx=−√^12 ,0≤−a( 2 √^12 −√^12 )≤^14. This can happen only
ifa=0. Therefore,P(x)=x^4 −x^2 +^18 =^18 T 4 (x).
194.From the identity
x^3 +1
x^3=
(
x+1
x) 3
− 3
(
x+1
x)
,
it follows that
√
r+1
√
r= 63 − 3 × 6 = 198.
Hence
(
√ (^4) r− 1
√ (^4) r
) 2
= 198 − 2 ,
and the maximum value of^4
√
r−√ (^41) ris 14.
(University of Wisconsin at Whitewater Math Meet, 2003, proposed by T. Andreescu)
195.Letx 1 =2 cosα,x 2 =2 cos 2α,..., xn =2 cosnα. We are to show that the
determinant
∣
∣∣
∣∣
∣∣
∣∣
T 0 (x 1 )T 0 (x 2 ) ··· T 0 (xn)
T 1 (x 1 )T 1 (x 2 ) ··· T 1 (xn)
..
.
..
.
... ..
.
Tn− 1 (x 1 )Tn− 1 (x 2 )···Tn− 1 (xn)∣
∣∣
∣∣
∣∣
∣∣
is nonzero. SubstitutingT 0 (xi)=1,T 1 (xi)=x,i= 1 , 2 ,...,n, and performing row
operations to eliminate powers ofxi, we can transform the determinant into
2 · 4 ··· 2 n−^1∣
∣∣
∣∣
∣∣
∣
∣
11 ··· 1
x 1 x 2 ··· xn
..
...
.
... ..
.
x 1 n−^1 xn 2 −^1 ···xnn−^1∣
∣∣
∣∣
∣∣
∣
∣
.
This is a Vandermonde determinant, and the latter is not zero sincexi =xj, for 1≤i<
j≤n, whence the original matrix is invertible. Its determinant is equal to