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Question Number 96287 by Ar Brandon last updated on 31/May/20

$$\mathcal{C}\mathrm{onsider}\mathrm{the}\mathrm{system}\mathrm{in}{\mathbb{N}}^3$$$$\left(\mathrm{S}\right):\{\begin{array}{l}{\mathrm{p}}^2+{\mathrm{q}}^2={\mathrm{r}}^2\\ \mathrm{q}+\mathrm{p}+\mathrm{r}=24\\ \mathrm{r}<\mathrm{p}+\mathrm{q}\end{array}$$$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{triplet}(\mathrm{p}:\mathrm{q}:\mathrm{r})\mathrm{is}\mathrm{solution}\mathrm{to}\left(\mathrm{S}\right)\mathrm{if}$$$$\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{r}<12.\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{are}\mathrm{solutions}\mathrm{to}\mathrm{the}\mathrm{equation};$$$${\mathrm{n}}^2-(24-\mathrm{r})\mathrm{n}+24(12-\mathrm{r})=0\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{an}\mathrm{unknown}.\mathrm{p}$$

Answered by maths mind last updated on 31/May/20

$${(p+q)}^2⩾p^2+q^2\Rightarrow p+q+r⩾2r\iff 24⩾2r\Rightarrow r⩽12$$$${(p+q)}^2={(24-r)}^2\iff p^2+q^2+2pq={(24-r)}^2$$$$\Rightarrow r^2+2pq=576-48r+r^2$$$$\Rightarrow pq=288-24r$$$$p+q=24-r\Rightarrow (p,q)/solutionof$$$$X^2-(24-r)X+288-24r=0$$$$\iff n^2-(24-r)n+12(24-2r)=0$$

Commented by Ar Brandon last updated on 31/May/20

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