Show that for all odd positive integer $n$, there exists a set $A$ where $A= [a_1, a_2, a_3, ... , a_n]$ and $\displaystyle\sum_{i=1}^n a_i =\prod_{i=1}^n a_i$.

Edit: $a_1,...,a_n$ must be distinct.

Here is my approach:

We can prove this by induction. The base case of $n = 1$ holds since $a_1 = a_1$. Now assume that there exists a set of $a_1,a_2,...,a_{2k-1}$ such that $a_1+a_2 +\cdots+a_{2k-1} = a_1a_2 \cdots a_{2k-1}$ is true for some $k$. If we add $a_{2k}+a_{2k+1}$ to both sides, we then must have numbers $a_2k$ and $a_{2k+1}$ such that $a_1a_2 \cdots a_{2k-1}+a_{2k}+a_{2k+1} = a_1a_2 \cdots a_{2k-1}a_{2k}a_{2k+1}$. We can think of this more simply as $x+y+z = xyz$ where $x = a_1a_2 \cdots a_{2k-1}, y = a_{2k}, z = a_{2k+1}$ and $y \neq z$ . We have that $z = \dfrac{x+y}{xy-1}$. We can keep increasing the $y$ value so that $y$ and $z$ are not equal to any of $a_1,a_2,...a_{2k-1}$ and $y \neq z$. Therefore we have finished the induction.

I am thinking my reasoning in the last step isn't very rigorous. Should I make it more rigorous?

  • $\begingroup$ Any restriction on $a_i$ such as distinct, positive, or integer? $\endgroup$ – Henry Dec 2 '15 at 17:41
  • $\begingroup$ Yes, I forgot to mention it must be distinct. $\endgroup$ – user19405892 Dec 2 '15 at 17:41
  • $\begingroup$ Do $a_i$ have to be integers? $\endgroup$ – fleablood Dec 2 '15 at 17:59
  • $\begingroup$ No they just must be distinct. $\endgroup$ – user19405892 Dec 2 '15 at 18:00
  • $\begingroup$ I think you need slightly more rigor on last step. a) show if (x + y)/(xy - 1) = (x + m)/(xm - 1) => y = m. Thus we can chose n + 1 y's such no y = a_i and therefore at least 1 z not equal to a_i. $\endgroup$ – fleablood Dec 2 '15 at 18:12

I think my answer isn't the one you wanted, but since there is no condition on $a_i$ except that they are distinct, for every $n$, you choose $A_n=\{m $ integer with $|m|\leq \lfloor n/2 \rfloor\}$. Since $0\in A$, the product is $0$ and since $m\in A_n \Rightarrow -m\in A_n$, the sum is also $0$.

  • $\begingroup$ That would hold under the condition that $n$ is even, not odd right since then we could pair every term with its negative? $\endgroup$ – user19405892 Dec 2 '15 at 17:57
  • $\begingroup$ @user19405892: No. An example would be $0,1,-1,2,-2$ with five terms, and zero product and sum. There is no equivalent of this type with an even number of terms $\endgroup$ – Henry Dec 2 '15 at 17:58
  • $\begingroup$ Oh, I see. Yes this is a great solution. It proves the question. But does my solution also work? $\endgroup$ – user19405892 Dec 2 '15 at 17:59

To answer your question directly:

Basic step: n = 1.

Than $a_1 = c$. (Any c; not just 1. H Potter gave an elegant answer for c = 0.)

Your induction step with a bit more rigor:

Let $x = \sum_{i=1}^{2k - 1}a_i = \prod_{i=1}^{2k - 1}a_i$

We want to find $y, z$ where $y \ne z; y \ne a_i; z \ne a_i$ and $x + y + z = xyz$.

So $z(xy - 1) = x + y$.

We can pick a y such that $y \ne 1/x$. (We can pick $y = 1/x$ if $x = - 1/x \ne a_i$ but that would mean $x^2 = -1$ so that isn't possible).

So $z = \frac {x+y}{xy - 1}$. But what if $z = a_i$? Well, we just pick another y; we have infinite choices. As $ \frac {x+y}{xy - 1} = \frac {x+y'}{xy' - 1} \implies (x +y)(xy' - 1) = (x + y')(xy - 1) \implies x^2(y' - y) = y - y' \implies x^2 = -1 OR y = y' \implies y = y'$, each different y will yield a different z. If we have really bad luck the first 2k - 1 choices of y will all yield z's equal some $a_i$, but the $2k$th choice will yield a distinct z.

Oh, we also need to pick $y$ so that $y \ne \frac {x+y}{xy - 1}$ (else z = y). So ... well, that's avoidable by simply not picking a $y$ that is a solution to $y^2 - 2y/x - x = 0$.

So, yes, we can always pick such a y and derive such a z.

[BTW: This argument does allow for x at some point equaling zero. There's no harm in selecting y = -x; z = 0. After that each $a_{2j+1} = - a_{2j}$.]


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