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Question Number 192350 by Abdullahrussell last updated on 15/May/23

Answered by Frix last updated on 15/May/23

$$\mathrm{Just}\mathrm{type}\mathrm{into}\mathrm{a}\mathrm{calculator}.$$$$\mathrm{Answer}\mathrm{is}\frac{5}{197}$$$$\mathrm{But}\mathrm{can}\mathrm{you}\mathrm{solve}\mathrm{this}?$$$$\frac{({(x-5)}^4+4)({(x-1)}^4+4)({(x+3)}^4+4)}{({(x-3)}^4+4)({(x+1)}^4+4)({(x+5)}^4+4)}$$

Commented by deleteduser1 last updated on 15/May/23

$$Usethefactthat{a}^4+4=[a(a-2)+2][a(a+2)+2]$$$$Take(x-5)(x-7)+2asthenumerator$$$$and(x+5)(x+7)+2asthedenominator$$$$whereanswer=\frac{Numerator}{Denominator}$$

Commented by Frix last updated on 15/May/23

$$\mathrm{Thank}\mathrm{you},\mathrm{I}\mathrm{already}\mathrm{know}\mathrm{this}.\mathrm{My}\mathrm{point}$$$$\mathrm{here}\mathrm{is},\mathrm{it}\mathrm{takes}\mathrm{too}\mathrm{long}\mathrm{to}\mathrm{get}\mathrm{the}‘‘{\mathrm{smart}}^{″}$$$$\mathrm{solution}\mathrm{when}\mathrm{using}\mathrm{a}\mathrm{calculator}\mathrm{is}10\mathrm{times}$$$$\mathrm{faster}.\mathrm{Only}\mathrm{the}\mathrm{general}\mathrm{case}\mathrm{is}\mathrm{interesting}.$$

Answered by deleteduser1 last updated on 15/May/23

$$a^4+4={(a^2+2)}^2-{\left(2a\right)}^2=(a^2-2a+2)(a^2+2a+2)$$$$=[a(a-2)+2][a(a+2)+2]$$$$\Rightarrow {(a+2)}^4+4=[(a+2)a+2][(a+2)(a+4)+2]$$$$\Rightarrow if{a}^4+4=pq,then,{(a+2)}^4=qr,{(a+4)}^4=rs,...$$$$(q>p;r>q;s>r;...)$$$$\Rightarrow \frac{(3^4+4)(7^4+4)({11}^4+4)}{(5^4+4)(9^4+4)({13}^4+4)}=\frac{ab\times cd\times ef}{bc\times de\times fg}=\frac{a}{g}$$$$3^4+4=5\times 17=ab\Rightarrow a=5;{13}^4+4=145\times 197=fg$$$$\Rightarrow \frac{a}{g}=\frac{5}{197}$$$$Generalsolution:$$$$\frac{(a^4+4)[{(a+4)}^4+4]...[{(a+4n)}^4+4]}{[{(a+2)}^4+4][{(a+6)}^4+4]...[{(a+4n+2)}^4+4]}$$$$=\frac{a(a-2)+2}{(a+4n+2)(a+4n+4)+2}$$

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