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Given three unknown positive integers. Is it possible to find the three numbers if we are given their Sum->(a+b+c) = X Product-> (abc) = Y Sum of Squares-> (a^2 + b^2 + c^2) = Z

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2 Answers 2

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Here is a method you can try. Let

$s_1=a+b+c$,

$s_2=a^2+b^2+c^2$,

$p_2=ab+bc+ac$,

$p_3=abc$

$a, b, c$ are the roots of the cubic $$0=(x-a)(x-b)(x-c)=x^3-s_1x^2+p_2x-p_3$$

We don't know $p_2$ but can calculate it using $s_1^2=s_2+2p_2$, and then solve the cubic to find $a,b,c$.

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  • $\begingroup$ Is there an easy method to calculate the roots if coefficients are integers..?? Or any method other than Hit-and-Trial. $\endgroup$
    – user143544
    Apr 17, 2014 at 14:15
  • $\begingroup$ @user143544 You need to find three different numbers, and one of the conditions you have fixes the third degree expression $abc$. So you almost inevitably have to solve a cubic (special cases excepted). Integer solutions are constrained (if coefficients are integers they must be factors of $p_3$). There are formulae and methods for solving the general cubic - e.g. Cardano method mathsa.ucd.ie/courses/mst3022/history4.pdf $\endgroup$ Apr 17, 2014 at 14:30
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Yes. Try squaring $X=(a+b+c)$.

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  • $\begingroup$ when is the product used ? $\endgroup$
    – T_O
    Apr 16, 2014 at 11:25
  • $\begingroup$ but then how to get 2(ab+bc+ca)..?? $\endgroup$
    – user143544
    Apr 16, 2014 at 11:26
  • $\begingroup$ It appears that I made a mistake whilst calculating as usual. Mark's method is best I can think of right now $\endgroup$
    – Jack Yoon
    Apr 16, 2014 at 11:36

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