Something something Newton's Sums

Algebra Level 4

{ x 1 7 + x 2 7 + x 3 7 + + x n 7 = 7 x 1 8 + x 2 8 + x 3 8 + + x n 8 = 8 x 1 9 + x 2 9 + x 3 9 + + x n 9 = 9 x 1 10 + x 2 10 + x 3 10 + + x n 10 = 10 \begin{cases} x_1^7+x_2^7+x_3^7+\cdots+x_n^7 =7\\ x_1^8+x_2^8+x_3^8+\cdots+x_n^8 =8\\ x_1^9+x_2^9+x_3^9+\cdots+x_n^9 =9\\ x_1^{10}+x_2^{10}+x_3^{10}+\cdots+x_n^{10} =10 \end{cases}

Does there exist positive real numbers x 1 , x 2 , x 3 , , x n x_1, x_2, x_3, \ldots, x_n with n 4 n\ge 4 such that the system of equations above is satisfied?

Yes, for all values of n n Yes, for finitely many values of n n Yes, for infinitely many values of n , n, but not all of them No

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1 solution

Steven Yuan
Jun 26, 2017

Assume there does exist positive real number solutions to the equations. Let S p = i = 0 n x i p , S_p = \displaystyle \sum_{i = 0}^n x_i^p, where p p is a positive integer. By the Cauchy-Schwarz Inequality,

S d S d + 2 = ( x 1 d + x 2 d + + x n d ) ( x 1 d + 2 + x 2 d + 2 + + x n d + 2 ) ( x 1 2 d + 2 + x 2 2 d + 2 + + x n 2 d + 2 ) 2 = ( x 1 d + 1 + x 2 d + 1 + + x n d + 1 ) 2 = S d + 1 2 . \begin{aligned} S_{d}S_{d + 2} &= (x_1^d + x_2^d + \dots + x_n^d)(x_1^{d + 2} + x_2^{d + 2} + \dots + x_n^{d + 2}) \\ &\geq \left ( \sqrt{x_1^{2d + 2}} + \sqrt{x_2^{2d + 2}} + \dots + \sqrt{x_n^{2d +2}} \right )^2 \\ &= (x_1^{d + 1} + x_2^{d + 1} + \dots + x_n^{d + 1})^2 \\ &= S_{d + 1}^2. \end{aligned}

Thus, S 7 S 9 S 8 2 , S_7S_9 \geq S_8^2, and S 8 S 10 S 9 2 . S_8S_{10} \geq S_9^2. Multiplying these together gives S 7 S 10 S 8 S 9 . S_7S_{10} \geq S_8S_9. We know that S 7 = 7 , S 8 = 8 , S 9 = 9 , S 10 = 10. S_7 = 7, S_8 = 8, S_9 = 9, S_{10} = 10. Plugging these in gives 7 ( 10 ) 8 ( 9 ) , 7(10) \geq 8(9), which is clearly false. Thus, there are no positive real numbers x 1 , x 2 , , x n x_1, x_2, \dots, x_n such that the system of equations is true.

Note that we can extend the argument to prove that, for all positive integers a , b a, b such that b a + 2 , b \geq a + 2, we have S a S b S a + 1 S b 1 . S_aS_b \geq S_{a + 1}S_{b - 1}.

Note: The "extended argument" is a special case of Majorization .

Calvin Lin Staff - 3 years, 11 months ago

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