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Consider an $n \times n$ matrix of $2n$ distinct numbers $a_1,a_2, \dots, a_n, b_1, b_2, \dots, b_n$, such that the number at the intersection of the $i$th row and the $j$th column is $a_i + b_j$:

$$ \left( {\begin{array}{cc} {a_1 + b_1} & a_1 + b_2 & \dots & a_1 + b_n\\ a_2 + b_1 & a_2 + b_2 & \dots & a_2 + b_n\\ . & . & & .\\ . & . & & .\\ . & . & & .\\ a_n + b_1 & a_n + b_2 & \dots & a_n + b_n \end{array} } \right). $$

Prove that if the product of numbers of each column is the same, then the product of numbers of each row is the same as well.

The statement of the question implies,

$$(b_i + a_1)(b_i + a_2)\dots(b_i + a_n) = (b_j + a_1)(b_j + a_2)\dots(b_j + a_n)$$

Let $\sigma_k$ denote the $k$th symmetric sum of $-a_1, -a_2, \dots, -a_n$. Therefore, the above simplifies to:

$$b_i^n - \sigma_1b_i^{n-1} + \sigma_2b_i^{n-2} + \dots + (-1)^n\sigma_n = b_j^n - \sigma_1b_j^{n-1} + \sigma_2b_j^{n-2} + \dots + (-1)^n\sigma_n$$

$$\implies (b_i^n - b_j^n) - \sigma_1(b_i^{n-1} - b_j^{n-1}) + \sigma_2(b_i^{n-2} - b_j^{n-2}) + \dots + (-1)^{n-1}\sigma_{n-1}(b_i - b_j) = 0$$

I have no idea how to proceed.

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  • $\begingroup$ Yes. I cannot see where you would apply calculus or linear algebra. In fact, the solution is algebraic. $\endgroup$
    – Gerard
    Nov 20 '13 at 10:34
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Let $f(x)=(x+a_1)\ldots (x+a_n),g(x)=(x+b_1)\ldots (x+b_n)$. Then $f(-a_i)=0$, and $f(b_i)=c$ don't depend on $i$. Denote $h(x)=(-1)^n(f(-x)-c)$. Since $h(-b_i)=0$, then $g=h$. Hence $(a_i+b_1)\ldots (a_i+b_n)=g(a_i)=-(-1)^nc=g(a_j)=(a_j+b_1)\ldots (a_j+b_n)$.

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