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Topic: A Sum And Product Puzzle (Read 725 times) |
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K Sengupta
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A Sum And Product Puzzle
« on: Feb 21st, 2007, 7:40am » |
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Let Sr = xr + yr + zr where x, y and z are real valued such that:
(i) S1 = 0; and
(ii) Sm+n/ (m+n) = (Sm/m) *(Sn/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
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| « Last Edit: Feb 22nd, 2007, 6:24am by K Sengupta » |
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Eigenray
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Re: A Sum And Product Puzzle
« Reply #1 on: Mar 8th, 2007, 3:09pm » |
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This isn't all that hard, but to keep it from being buried:| hidden: |
Let Sn(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 Sm+n(x) = (m+n) Sm(x) Sn(x), (*)
with equality as polynomials in x. Now, if n is odd, then
Sn(x) = xn + 1 - (x+1)n
= -n xn-1 - C(n,2) xn-2 - ... - nx,
while if n is even, then
Sn(x) = xn + 1 + (x+1)n
= 2xn + n xn-1 + ... + n x + 2.
Case 1: m,n both even. Then (*) is
mn (2xm+n + ...) = (m+n) (2xm+...)(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) xm+n-1 + ... ] = (m+n) [2xm + ...][ nxn-1 + ...],
and by comparing coefficients of xm+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)xn-1 + ... ] = (n+2) [2x2 + 2x + 2][ nxn-1 + C(n,2)xn-2 + C(n,3)xn-3 + ...],
so comparing coefficients of xn-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 Sm, Sn have degrees m-1, n-1, respectively, while Sm+n has degree m+n, so (*) can't hold. |
So there are no solutions other than the ones given.
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