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let $a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ be real numbers,and such $a_{i}\neq a_{j},\forall i\neq j$.

Assmue that: there exsit real number $\alpha$ such for every $i=1,2,3,\cdots,n$ have $$(a_{i}+b_{1})(a_{i}+b_{2})(a_{i}+b_{3})\cdots (a_{i}+b_{n})=\alpha$$

show that: there must exsit real number $\beta$ such for any $j=1,2,3,\cdots,n$ such $$(a_{1}+b_{j})(a_{2}+b_{j})\cdots (a_{n}+b_{j})=\beta$$

I think this is very interesting ,Fell it's clear,But these problem I can't How explain this is why?

My idea: if $\alpha$ such $$\sum_{k=1}^{n}\ln{|(a_{i}+b_{k})|}=\ln{|\alpha|},i=1,2,\cdots,n$$ then we must choice a real number $\beta$ such $$\sum_{k=1}^{n}\ln{|(a_{k}+b_{j})|}=\ln{|\beta|},j=1,2,\cdots,n$$ Maybe we can use $\ln{n}$ is dense? and I think this problem condition $a_{i}\neq a_{j},\forall i\neq j$ is important? maybe we always use contradiction?

Thank you

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  • $\begingroup$ "My idea:" ?? Why taking logarithms would change anything? "Maybe we can use lnn is dense?" ?? Not dense, and does not appear in the question. $\endgroup$ – Did Nov 29 '14 at 17:57
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Put $\displaystyle P(x)=\prod_{k=1}^n (x+b_k)$. Then $P$ is a degree $n$ polynomial, with leading coefficient $1$. Your hypothesis say that $P(x)-\alpha$ has the $n$ distincts roots $a_i$, $i=1,\cdots,n$. Hence $P(x)-\alpha=\prod_{i=1}^n (x-a_i)$ (taking in account the leading coefficient). Now replace $x$ by $-x$, we get $$\prod_{i=1}^n (x+a_i)=(-1)^{n-1}\alpha+\prod_{j=1}^n (x-b_j)$$ and it suffices to replace $x$ by $b_j$ to finish.

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