If I have $ n $-positive integers, and I compute their sum and product, is there any different group of $ n $-positive integers that will have the same sum and product?

For example, if $ a,...,z $ denote 26 positive integers, and we define:

\begin{align} a+b+c+d+....+z &= \text{Sum} \\ a \cdot b \cdot c \cdot d \cdot .... \cdot z &= \text{Product} \end{align}

Is there any way I can get the same Sum and Product from a different group of 26 (in this example) positive integers?

A friend of mine pointed out that knowing that we have a group of 3 that works, we can show that it works for all positive groups of $ n $ integers.

For Example: $ \{3,3,10 \} $ and $ \{2,5,9 \} $ both yield Sum $=16 $ and Product $=90 $.
Now we can just continually add a number (let's say 1) as the next integer to get multiple solutions for $ n =4,5,6,... $.
Explicitly, $ \{ 3,3,10,1 \} $ and $ \{ 2,5,9,1 \} $ both give Sum$=17$ and Product$=90$.

  • $\begingroup$ @dxiv Thank you for the point out, I think I have misinterpreted this question. I will delete the above comment. $\endgroup$ – Teresa Lisbon Sep 26 '16 at 5:04
  • 2
    $\begingroup$ Two sets of 3 positive integers with equal sum and product has explicit examples of different sets that have the same sum and product for $n=3$. $\endgroup$ – dxiv Sep 26 '16 at 5:12
  • 1
    $\begingroup$ Here is one: $\{ 2,5,9\}$ and $\{ 3,3,10\}$. Here's another : $\{ 2,6,6\}, \{ 3, 3, 8\}$. Here's yet another: $\{ 2,7,12\}$ and $\{ 3,4,14\}$. Another: $\{ 2,8,30\}$ and $\{ 3,5,32\}$. WOW, I never thought this question was so good! $+1$, I wish I could give more. $\endgroup$ – Teresa Lisbon Sep 26 '16 at 5:15
  • $\begingroup$ That's a good question. Don't think so. I think nm $\ne$ (n-k)(m+k) = nm + k (n-m) -k^2 is provable by am/gm and the rest by induction. I might be wrong though. ... yeah k (n-m) - k^2 iff k = 0 or n-m =k which means either n=m or nm=0. Both assumed not true. So I leave induction up to you. (Might not be true) $\endgroup$ – fleablood Sep 26 '16 at 5:18

Given any of lots of pairs $\{a,b\}, \{d,e\}$, you can add a number to each to make matching sum and product. We want $$a+b+c=d+e+f\\abc=def\\f-c=a+b-d-e\\\frac fc = \frac {ab}{de}$$ and we can solve the last two to get $$c=(a+b-d-e)\frac 1{\frac {ab}{de}-1}\\f=(d+e-a-b)\frac 1{\frac {de}{ab}-1}$$ As long as $a+b \neq d+e$ and none are zero we have a solution. As long as the pair with the greater sum also has the greater product both $c,f$ will be positive.

| cite | improve this answer | |

Well it can't be true for 2 values.

$n >0,m>0$ then $nm>0$.

Let $(n+k)+(m-k) = n+m $ and $(n+k)(m-k)=nm $

Then $nm +k (m-n) - k^2 = nm $ so

$k (m-n)= k^2$. If $k \ne 0$ then $m-n=k $ so $m-k = n $ and $n+k=m $.

If $k =0$ then $n+k=n $ and $m+k=m$.

So $n,m $ are the only positive integers (indeed positive reals) where $a+b=n+m;ab=nm$.

This shouldn't surprise us. It's just a restatement of the arithmetic mean vs geometric mean property.

Can we extend that via induction?

To be honest I'm not sure but I think we can.

Or maybe directly

$(a_n - \sum k_i)\prod(a_i + k_i)=a_n\prod a_i \implies \{a_i +k_i,a_n - \sum k_i\} = \{a_i\} $

Can that be proven

| cite | improve this answer | |
  • $\begingroup$ For 2 values, we can't find any other positive integers with the same sum and product, as you mention. However, looks like for 3 or more we can (by modifying the script by @ShreevatsaR pointed to from @dxiv). So can we find multiple solutions for all n>2 then? If so, why? $\endgroup$ – ItM Sep 26 '16 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.