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Question Number 27996 by JI Siam last updated on 18/Jan/18

$$\mathrm{p},\mathrm{q}\mathrm{are}\mathrm{two}\mathrm{natural}\mathrm{number}\mathrm{and}$$$$\frac{{\mathrm{p}}^6+{2\mathrm{p}}^4+{4\mathrm{p}}^2}{{\mathrm{p}}^9-{8\mathrm{p}}^3}-\frac{1}{4\mathrm{q}}=\frac{5}{6\mathrm{q}},$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{minimum}$$$$\mathrm{possible}\mathrm{value}\mathrm{of}\mathrm{p}+\mathrm{q}$$

Answered by prakash jain last updated on 18/Jan/18

$$\frac{{\mathrm{p}}^6+{2\mathrm{p}}^4+{4\mathrm{p}}^2}{{\mathrm{p}}^9-{8\mathrm{p}}^3}-\frac{1}{4\mathrm{q}}=\frac{5}{6\mathrm{q}},$$$$\frac{p^2(p^4+{2\mathrm{p}}^2+4)}{p^3(p^6-2^3)}=\frac{5}{6q}+\frac{1}{4q}$$$$\frac{(p^4+{2\mathrm{p}}^2+4)}{p(p^2-2)(p^4+2p^2+4)}=\frac{13}{12q}$$$$q=\frac{13}{12}p(p^2-2)$$$$p(p^2-2)\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}12$$$$p=6\mathrm{q}=\frac{13}{2}(6^2-2)$$

Commented by prakash jain last updated on 18/Jan/18

$$\mathrm{p}=13,\mathrm{q}=12({13}^2-2)$$

Commented by JI Siam last updated on 18/Jan/18

$$\mathrm{i}\mathrm{think}\mathrm{p}=12,\mathrm{q}=13({12}^2-2)\mathrm{sir}$$

Commented by prakash jain last updated on 18/Jan/18

$$\mathrm{thanks}.\mathrm{corrected}\mathrm{p}=6$$

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