Algebra
81.Assume that both numbers are perfect cubes. Then so is their product
(n+ 3 )(n^2 + 3 n+ 3 )=n^3 + 6 n^2 + 12 n+ 9.
However, this number differs from the perfect cube(n+ 2 )^3 =n^3 + 6 n^2 + 12 n+8by
one unit. And this is impossible because no perfect cubes can be consecutive integers
(unless one of them is zero). This proves the claim.
82.Letm=pq. We use the identity
xm−ym=(x−y)(xm−^1 +xm−^2 y+···+ym−^1 ),
which can be applied to the matricesAand−Bsince they commute. We have
(A−(−B))(Am−^1 +Am−^2 (−B)+···+(−B)m−^1 )
=Am−(−B)m=(Ap)q−(− 1 )pq(Bq)p=In.
Hence the inverse ofA+B=A−(−B)isAm−^1 +Am−^2 (−B)+···+(−B)m−^1.
83.First solution: LetF(x)be the polynomial in question. IfF(x)is the square of a
polynomial, then writeF(x)=G(x)^2 + 02. In general,F(x)is nonnegative for all real
numbersxif and only if it has even degree and is of the form
F(x)=R(x)^2 (x^2 +a 1 x+b 1 )(x^2 +a 2 x+b 2 )···(x^2 +anx+bn),
where the discriminant of each quadratic factor is negative. Completing the square
x^2 +akx+bk=
(
x+
ak
2
) 2
+ 2 , with =
√
bk−
ak^2
4
,
we can write
