System of $24$ variables

Question

Assume that $a_1, a_2,\ldots, a_{24}$ satisfy $$a_1+a_2+\ldots+a_{24}=26$$$$a_1^2+a_2^2+\ldots+a_{24}^2=26$$$$\vdots$$$$a_1^{24}+a_2^{24}+\ldots+a_{24}^{24}=26$$ Find $a_1a_2⋯a_{24}$. How do I solve it? I know there is sone thing I am missing, nobody will solve the equations.

Answer 1

take $$f(x)=(x-a_{1})(x-a_{2})\cdots (x-a_{24})=x^n-\sigma_{1}x^{n-1}+\sigma_{2}x^{n-2}-\cdots+\sigma_{24}$$ let $$s_{k}=a^k_{1}+\cdots+a^k_{24}\Longrightarrow s_{k}=24,k=1,2,3\cdots,24$$ and use well identity: $$S_{n}-\sigma_{1}S_{n-1}+\cdots+24\sigma_{24}=0$$ so $$1-\sigma_{1}+\sigma_{2}+\cdots+\sigma_{24}=0$$ so we have $f(1)=0$. WLOG,Assmue that $a_{24}=1$

and use same methods,we have $a_{23}=a_{23}=\cdots=a_{1}=1$

Answer 2

Let $$S_k(a_1,\ldots,a_n)=\sum_{j=1}^{n} a_j^k$$ be the power sums and $\Pi_k(a_1,\ldots,a_n)$ be the homogeneous symmetric polynomials in $a_1,\ldots,a_n$ having degree $k$. By the Newton-Girard formulas we have: $$ S_1=\Pi_1,$$ $$ S_2 - S_1 \Pi_1= -2\Pi_2, $$ $$ S_3 - S_2 \Pi_1 + S_1\Pi_2 = 3\Pi_3, $$ $$\ldots $$ $$ S_{24}-S_{23}\Pi_1 +\ldots -S_1\Pi_{23} =-24\Pi_{24},$$ and we want to compute $\Pi_{24}$ from $S_1=S_2=\ldots=S_{24}=26$. That leads to:

$$-24\Pi_{24} = 26\left(1-\Pi_1+\Pi_2-\ldots-\Pi_{23}\right)=23\Pi_{23}-26\Pi_{23} $$ hence: $$ \Pi_{24}=\frac{1}{8}\Pi_{23} $$ and going on: $$ 23\Pi_{23} = 26(1-\Pi_1+\Pi_2-\ldots+\Pi_{22}) = -22\Pi_{22}+26\Pi_{22}$$ hence: $$\Pi_{23}=\frac{4}{23}\Pi_{22} $$ and continuing this way we get: $$ \Pi_{24} = \frac{3}{24}\cdot\frac{4}{23}\cdot\frac{5}{22}\Pi_{21} = \frac{3\cdot 4\cdot 5\cdot\ldots\cdot 25}{24!}\Pi_1=25\cdot 13 = \color{red}{325}.$$