p-q-are-two-natural-number-and-p-6-2p-4-4p-2-p-9-8p-3-1-4q-5-6q-then-find-the-minimum-possible-value-of-p-q-

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

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

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

((p^6 +2p^4 +4p^2 )/(p^9 −8p^3 ))−(1/(4q))=(5/(6q)), ((p^2 (p^4 +2p^2 +4))/(p^3 (p^6 −2^3 )))=(5/(6q))+(1/(4q)) (((p^4 +2p^2 +4))/(p(p^2 −2)(p^4 +2p^2 +4)))=((13)/(12q)) q=((13)/(12))p(p^2 −2) p(p^2 −2) is a multiple of 12 p=6 q=((13)/2)(6^2 −2)

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

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