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?

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- Nov 16th 2011, 11:23 AMLasombrasums 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?

- Nov 16th 2011, 06:06 PMMonroeYoderRe: sums and products...
For 2 different pairs of natural numbers we can do it without much difficulty. Let our pairs be $\displaystyle \{a,b\}$ and $\displaystyle \{c,d\}$ and remember these sets are not equal.

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

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

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

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

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

Therefore $\displaystyle \{a,b\}=\{c,d\}$ but this is a contradiction. So two different pairs of natural numbers cannot have both the same sum and same product. - Nov 17th 2011, 03:36 AMOpalgRe: sums and products...