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Topic: Sum The Odd Powers, Get Real Triplets (Read 336 times) |
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K Sengupta
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Sum The Odd Powers, Get Real Triplets
« on: Jul 27th, 2007, 8:10am » |
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Analytically determine all possible real triplets (p, q, r) satisfying:
p+q+r = 0
P3 + q3 + r3 = 18
p7 + q7 + r7 = 2058
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Eigenray
wu::riddles Moderator
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Re: Sum The Odd Powers, Get Real Triplets
« Reply #1 on: Aug 5th, 2007, 3:34am » |
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| hidden: | 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
xk = axk-2 + bxk-3
when x = p,q, or r, and so if sk = pk + qk + rk, then
sk = ask-2 + bk-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 =  7. So {p,q,r} satisfy the cubic x3 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}. |
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| « Last Edit: Aug 5th, 2007, 3:37am by Eigenray » |
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