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        riddles >> hard >> A Sum And Product Puzzle
        
(Message started by: K Sengupta on Feb 21^st, 2007, 7:40am)

Title: A Sum And Product Puzzle
Post by K Sengupta on Feb 21^st, 2007, 7:40am

Let S_r = x^r+ y^r + z^r where x, y and z are real valued such that:

(i) S₁ = 0; and

(ii) S_m+n/ (m+n) = (S_m/m) *(S_n/n) for (m,n) = (2,3); (3,2); (2,5); (5,2)

Determine all other pairs (m,n) such that both (i) and (ii) holds for all real numbers x, y and z such that x+ y+ z = 0

Title: Re: A Sum And Product Puzzle
Post by Eigenray on Mar 8^th, 2007, 3:09pm

This isn't all that hard, but to keep it from being buried:[hideb]
Let S_n(x) = xⁿ + 1 + (-x-1)ⁿ. Then (ii) holds for all x,y,z such that x+y+z=0 if and only if

mn S_m+n(x) = (m+n) S_m(x) S_n(x), (*)

with equality as polynomials in x. Now, if n is odd, then

S_n(x) = xⁿ + 1 - (x+1)ⁿ
= -n x^n-1 - C(n,2) x^n-2 - ... - nx,

while if n is even, then

S_n(x) = xⁿ + 1 + (x+1)ⁿ
= 2xⁿ + n x^n-1 + ... + n x + 2.

Case 1: m,n both even. Then (*) is

mn (2x^m+n + ...) = (m+n) (2x^m+...)(2xⁿ+...),

so we need 2mn = 4(m+n), or (2m-2)(2n-2) = 4, and since m,n are even, this forces m=n=2. Then we can check that

(2+2)S₂(x)S₂(x) = 16x⁴ + 32x³ + 48x² + 32x + 16 = 8 S₄(x),

so (2,2) is not a solution.

Case 2: m,n different parity; WLOG m even, n odd. Then (*) becomes

mn [(m+n) x^m+n-1 + ... ] = (m+n) [2x^m + ...][ nx^n-1 + ...],

and by comparing coefficients of x^m+n-1, we need mn = 2n, which requires m=2, and so, assuming n>3, we have

2n [(n+2) xⁿ⁺¹ + C(n+2,2)xⁿ + C(n+2,3)x^n-1 + ... ] = (n+2) [2x² + 2x + 2][ nx^n-1 + C(n,2)x^n-2 + C(n,3)x^n-3 + ...],

so comparing coefficients of x^n-1 gives

2n C(n+2,3) = (n+2)[ 2*C(n,3) + 2*C(n,2) + 2*n ],
2n(n+2)(n+1)n/6 = (n+2)2[ n(n-1)(n-2)/6 + n(n-1)/2 + n ]
= 2(n+2)n [ n² + 5 ]/6,

or n(n+1) = n²+5, i.e., n=5, and we can check that (2,5) is a solution. Since the above assumed n>3, we need to check (2,3), which is also a solution.

Case 3: m,n both odd. Then S_m, S_n have degrees m-1, n-1, respectively, while S_m+n has degree m+n, so (*) can't hold.[/hideb]

So there are no solutions other than the ones given.