Question Number 192350 by Abdullahrussell last updated on 15/May/23
Answered by Frix last updated on 15/May/23
$$\mathrm{Just}\:\mathrm{type}\:\mathrm{into}\:\mathrm{a}\:\mathrm{calculator}. \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{197}} \\ $$$$\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\frac{\left(\left({x}−\mathrm{5}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}−\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}+\mathrm{3}\right)^{\mathrm{4}} +\mathrm{4}\right)}{\left(\left({x}−\mathrm{3}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}+\mathrm{5}\right)^{\mathrm{4}} +\mathrm{4}\right)} \\ $$
Commented by deleteduser1 last updated on 15/May/23
![Use the fact that a^4 +4=[a(a−2)+2][a(a+2)+2] Take (x−5)(x−7)+2 as the numerator and (x+5)(x+7)+2 as the denominator where answer=((Numerator)/(Denominator))](Q192365.png)
$${Use}\:{the}\:{fact}\:{that}\:{a}^{\mathrm{4}} +\mathrm{4}=\left[{a}\left({a}−\mathrm{2}\right)+\mathrm{2}\right]\left[{a}\left({a}+\mathrm{2}\right)+\mathrm{2}\right] \\ $$$${Take}\:\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right)+\mathrm{2}\:{as}\:{the}\:{numerator} \\ $$$${and}\:\left({x}+\mathrm{5}\right)\left({x}+\mathrm{7}\right)+\mathrm{2}\:{as}\:{the}\:{denominator} \\ $$$${where}\:{answer}=\frac{{Numerator}}{{Denominator}} \\ $$
Commented by Frix last updated on 15/May/23
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{I}\:\mathrm{already}\:\mathrm{know}\:\mathrm{this}.\:\mathrm{My}\:\mathrm{point} \\ $$$$\mathrm{here}\:\mathrm{is},\:\mathrm{it}\:\mathrm{takes}\:\mathrm{too}\:\mathrm{long}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:``\mathrm{smart}'' \\ $$$$\mathrm{solution}\:\mathrm{when}\:\mathrm{using}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{is}\:\mathrm{10}\:\mathrm{times} \\ $$$$\mathrm{faster}.\:\mathrm{Only}\:\mathrm{the}\:\mathrm{general}\:\mathrm{case}\:\mathrm{is}\:\mathrm{interesting}. \\ $$
Answered by deleteduser1 last updated on 15/May/23
![a^4 +4=(a^2 +2)^2 −(2a)^2 =(a^2 −2a+2)(a^2 +2a+2) =[a(a−2)+2][a(a+2)+2] ⇒(a+2)^4 +4=[(a+2)a+2][(a+2)(a+4)+2] ⇒if a^4 +4=pq,then,(a+2)^4 =qr,(a+4)^4 =rs,... (q>p;r>q;s>r;...) ⇒(((3^4 +4)(7^4 +4)(11^4 +4))/((5^4 +4)(9^4 +4)(13^4 +4)))=((ab×cd×ef)/(bc×de×fg))=(a/g) 3^4 +4=5×17=ab⇒a=5;13^4 +4=145×197=fg ⇒(a/g)=(5/(197)) General solution: (((a^4 +4)[(a+4)^4 +4]...[(a+4n)^4 +4])/([(a+2)^4 +4][(a+6)^4 +4]...[(a+4n+2)^4 +4])) =((a(a−2)+2)/((a+4n+2)(a+4n+4)+2))](Q192364.png)
$${a}^{\mathrm{4}} +\mathrm{4}=\left({a}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{2}\right)\left({a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{2}\right) \\ $$$$=\left[{a}\left({a}−\mathrm{2}\right)+\mathrm{2}\right]\left[{a}\left({a}+\mathrm{2}\right)+\mathrm{2}\right] \\ $$$$\Rightarrow\left({a}+\mathrm{2}\right)^{\mathrm{4}} +\mathrm{4}=\left[\left({a}+\mathrm{2}\right){a}+\mathrm{2}\right]\left[\left({a}+\mathrm{2}\right)\left({a}+\mathrm{4}\right)+\mathrm{2}\right] \\ $$$$\Rightarrow{if}\:\:{a}^{\mathrm{4}} +\mathrm{4}={pq},{then},\left({a}+\mathrm{2}\right)^{\mathrm{4}} ={qr},\left({a}+\mathrm{4}\right)^{\mathrm{4}} ={rs},... \\ $$$$\left({q}>{p};{r}>{q};{s}>{r};...\right) \\ $$$$\Rightarrow\frac{\left(\mathrm{3}^{\mathrm{4}} +\mathrm{4}\right)\left(\mathrm{7}^{\mathrm{4}} +\mathrm{4}\right)\left(\mathrm{11}^{\mathrm{4}} +\mathrm{4}\right)}{\left(\mathrm{5}^{\mathrm{4}} +\mathrm{4}\right)\left(\mathrm{9}^{\mathrm{4}} +\mathrm{4}\right)\left(\mathrm{13}^{\mathrm{4}} +\mathrm{4}\right)}=\frac{{ab}×{cd}×{ef}}{{bc}×{de}×{fg}}=\frac{{a}}{{g}} \\ $$$$\mathrm{3}^{\mathrm{4}} +\mathrm{4}=\mathrm{5}×\mathrm{17}={ab}\Rightarrow{a}=\mathrm{5};\mathrm{13}^{\mathrm{4}} +\mathrm{4}=\mathrm{145}×\mathrm{197}={fg} \\ $$$$\Rightarrow\frac{{a}}{{g}}=\frac{\mathrm{5}}{\mathrm{197}} \\ $$$${General}\:{solution}: \\ $$$$\frac{\left({a}^{\mathrm{4}} +\mathrm{4}\right)\left[\left({a}+\mathrm{4}\right)^{\mathrm{4}} +\mathrm{4}\right]...\left[\left({a}+\mathrm{4}{n}\right)^{\mathrm{4}} +\mathrm{4}\right]}{\left[\left({a}+\mathrm{2}\right)^{\mathrm{4}} +\mathrm{4}\right]\left[\left({a}+\mathrm{6}\right)^{\mathrm{4}} +\mathrm{4}\right]...\left[\left({a}+\mathrm{4}{n}+\mathrm{2}\right)^{\mathrm{4}} +\mathrm{4}\right]} \\ $$$$=\frac{{a}\left({a}−\mathrm{2}\right)+\mathrm{2}}{\left({a}+\mathrm{4}{n}+\mathrm{2}\right)\left({a}+\mathrm{4}{n}+\mathrm{4}\right)+\mathrm{2}} \\ $$