MHB Finding Real Solutions for a System of Equations

[anemone]

Find all solutions in real numbers to the system below:

$a_1+a_2+\cdots+a_{1997}=1997$

$a_1^3+a_2^3+\cdots+a_{1997}^3=a_1^4+a_2^4+\cdots+a_{1997}^4$

 

[Opalg]

[sp]The only solution is $a_j = 1$ for all $j$ ($1\leqslant j\leqslant 1997$).

Write the equations as $$\sum_{j=1}^{1997}(a_j-1) = 0, \qquad (1)$$ $$\sum_{j=1}^{1997}a_j^3(a_j-1) = 0. \qquad(2)$$ Subtract (1) from (2): $$\sum_{j=1}^{1997}(a_j^3 - 1)(a_j-1) = 0. \qquad(3)$$ In each term of that sum, the numbers $a_j^3 - 1$ and $a_j - 1$ are either both positive, both zero or both negative, depending on whether $a_j \gtrless 1$. So each term is non-negative, and the sum can only be zero if each term is zero. Therefore $a_j = 1$ for all $j$.[/sp]

 

[anemone]

[sp]The only solution is $a_j = 1$ for all $j$ ($1\leqslant j\leqslant 1997$).

Write the equations as $$\sum_{j=1}^{1997}(a_j-1) = 0, \qquad (1)$$ $$\sum_{j=1}^{1997}a_j^3(a_j-1) = 0. \qquad(2)$$ Subtract (1) from (2): $$\sum_{j=1}^{1997}(a_j^3 - 1)(a_j-1) = 0. \qquad(3)$$ In each term of that sum, the numbers $a_j^3 - 1$ and $a_j - 1$ are either both positive, both zero or both negative, depending on whether $a_j \gtrless 1$. So each term is non-negative, and the sum can only be zero if each term is zero. Therefore $a_j = 1$ for all $j$.[/sp]

Well done Opalg and thanks for participating! Your solution is elegant and compelling!

Here is a solution that is proposed by other:

Spoiler

First, we let $S_n=a_1^n+a_2^n+\cdots+a_{1997}^n$.

By the power mean inequality, we have

$\left( \dfrac{S_4}{1997} \right)^{\dfrac{1}{4}}\ge \dfrac{S_1}{1997}=1$ and

$\left( \dfrac{S_4}{1997} \right)^{\dfrac{1}{4}}\ge \left(\dfrac{S_3}{1997} \right)^{\dfrac{1}{4}}=\left( \dfrac{S_4}{1997} \right)^{\dfrac{1}{3}}$

and so $\dfrac{S_4}{1997}\le 1$ as well.

Thus equality holds in the power mean inequality, which implies $a_1=a_2=\cdots=a_{1997}=1$.