362 Algebra
88.The solution is based on the identity
ak+bk=(a+b)(ak−^1 +bk−^1 )−ab(ak−^2 +bk−^2 ).This identity arises naturally from the fact that bothaandbare solutions to the equation
x^2 −(a+b)x+ab=0, hence also toxk−(a+b)xk−^1 +abxk−^2 =0.
Assume that the conclusion is false. Then for somen,a^2 n+b^2 nis divisible bya+b.
Fork = 2 n, we obtain that the right-hand side of the identity is divisible bya+b,
hence so isab(a^2 n−^2 +b^2 n−^2 ). Moreover,aandbare coprime toa+b, and therefore
a^2 n−^2 +b^2 n−^2 must be divisible bya+b. Through a backward induction, we obtain that
a^0 +b^0 =2 is divisible bya+b, which is impossible sincea, b >1. This contradiction
proves the claim.
(R. Gelca)
89.Letnbe an integer and letn
(^3) −n
6 =k. Becausen
(^3) −nis the product of three consecutive
integers,n− 1 ,n,n+1, it is divisible by 6; hencekis an integer. Then
n^3 −n= 6 k=(k− 1 )^3 +(k+ 1 )^3 −k^3 −k^3.
It follows that
n=n^3 −(k− 1 )^3 −(k+ 1 )^3 +k^3 +k^3 ,
and thus
n=n^3 +
(
1 −
n^3 −n
6) 3
+
(
− 1 −
n^3 +n
6) 3
+
(
n^3 −n
6) 3
+
(
n^3 −n
6) 3
.
Remark.Lagrange showed that every positive integer is a sum of at most four perfect
squares. Wieferich showed that every positive integer is a sum of at most nine perfect
cubes of positive integers. Waring conjectured that in general, for everynthere is a
numberw(n)such that every positive integer is the sum of at mostw(n) nth powers of
positive integers. This conjecture was proved by Hilbert.
90.First solution: Using the indentity
a^3 +b^3 +c^3 − 3 abc=1
2
(a+b+c)((a−b)^2 +(b−c)^2 +(c−a)^2 )applied to the (distinct) numbersa=^3
√
x−1,b=^3√
x, andc=^3√
x+1, we transform
the equation into the equivalent
(x− 1 )+x+(x+ 1 )− 33√
(x− 1 )x(x+ 1 )= 0.