1. ## sums and products...

i suppose this is the correct place to post this question...it is easy to prove that 2 different pairs of Natural numbers cannot have the same sum AND the same product.Can this be proven for groups of 3,4,5,...n Natural numbers?

2. ## Re: sums and products...

For 2 different pairs of natural numbers we can do it without much difficulty. Let our pairs be $\{a,b\}$ and $\{c,d\}$ and remember these sets are not equal.

Suppose that $ab=cd$ and $a+b=c+d$. Since $a\not = c$ we can say $a=c+k$ for some non-zero number $k$

To ensure $a+b=c+d$ we then know that $b=d-k$

Then $ab=(c+k)(d-k)=cd-ck+dk-k^2$ and since $ab=cd$ we know that we must have $-ck+dk-k^2=0$

So by factoring we have $k(d-c)-k^2 = k((d-c)-k)=0$. Remember that $k\not = 0$ and so $d-c-k=0$ which means $k=d-c$

Therefore $a=c+k=c+d-c=d$ and $b=d-k=d-(d-c)=c$

Therefore $\{a,b\}=\{c,d\}$ but this is a contradiction. So two different pairs of natural numbers cannot have both the same sum and same product.

3. ## Re: sums and products...

Originally Posted by Lasombra
i suppose this is the correct place to post this question...it is easy to prove that 2 different pairs of Natural numbers cannot have the same sum AND the same product.Can this be proven for groups of 3,4,5,...n Natural numbers?
It doesn't work for sets of four numbers. The sets {1,4,4,6} and {2,2,3,8} each have sum 15 and product 96.