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        riddles >> medium >> Sum The Odd Powers, Get Real Triplets
        
(Message started by: K Sengupta on Jul 27^th, 2007, 8:10am)

Title: Sum The Odd Powers, Get Real Triplets
Post by K Sengupta on Jul 27^th, 2007, 8:10am

Analytically determine all possible real triplets (p, q, r) satisfying:

p+q+r = 0
P³ + q³ + r³ = 18
p⁷ + q⁷ + r⁷ = 2058

Title: Re: Sum The Odd Powers, Get Real Triplets
Post by Eigenray on Aug 5^th, 2007, 3:34am

[hideb]Since p+q+r=0, we have

f(x) = (x-p)(x-q)(x-r) = x³ - ax - b,

where a = -(pq + qr + pr), and b = pqr. Thus

x^k = ax^k-2 + bx^k-3

when x = p,q, or r, and so if s_k = p^k + q^k + r^k, then

s_k = as_k-2 + b_k-3.

Also note that s₂ = s₁² + 2a = 2a. Now,

18 = s₃ = as₁ + b₀ = 3b

implies b=6, and then

2058 = s₇ = as₅ + 6s₄
= a(as₃ + 6s₂) + 6(as₂ + 6s₁)
= a(18a + 12a) + 6(2a²) = 42a²,

or a = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 7. So {p,q,r} satisfy the cubic x³ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 7x - 6 = 0. If a=-7, the discriminant is 4(7)³ + 27(-6)² > 0, so the roots are not all real. If a=7, we get x³ - 7x - 6 = (x+1)(x+2)(x-3), so {p,q,r} = {-1,-2,3}.[/hideb]