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I have an ordered pair $(ac+bd, ad+cb)$ where $a$, $b$, $c$ and $d$ are natural numbers. How can I rearrange that pair, so that exactly 3 multiplications, 3 additions and 1 subtraction are used?

I tried to place some of the factors outside the brackets for each component of the pair, but that always involves a division. I think that I have to add additional variables somehow, but I have no idea how to do that.

Can you give me a hint?

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    HINT Try using $(a+b) \cdot (c+d)$.2011-11-09
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    Thank you! But how can I apply this to the ordered pair? When I subtract the terms of the respective components of the pair like this: $((a+b)(c+d)-(ad+bc), (a+b)(c+d)-(ac+bd))$ the result would be correct, but that uses exactly twice as much operations as allowed.2011-11-09
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    @Zoidberg: you don't need to think of it as an ordered pair. You just need to produce the values of $ac+bd$ and $ad+cb$ with the given number of operations. So define $e=(a+b)(c+d)$ (which uses one multiply and two adds) and see if you can use the rest of the calculations given to get where you need to be. In particular, the sum of the two is exactly $e$, so if you can get one with the subtract left, you can do (ac+bd,e-(ac,bd))2011-11-09
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    @RossMillikan: Nice answer. The little typo may confuse the OP. So, to the OP you may wish to define $r=a.c+b.d$ and your answer is then ($r$,$e-r$).2011-11-09
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    @Swapan: Right you are. Thanks2011-11-09
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    Great! Thank you very much for your help, Ross, Srivatsan and Swapan!2011-11-10

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