Question Number 11049 by Joel576 last updated on 09/Mar/17
$$\mathrm{If}\:\:{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}{n}\:+\:\mathrm{10} \\ $$$$\mathrm{Find}\:\mathrm{out}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{for}\:{n}\: \\ $$
Commented by ajfour last updated on 09/Mar/17
$${cant}\:{figure}\:{any}\:{other}\:{than}\:\pm\mathrm{3}\:. \\ $$
Answered by mrW1 last updated on 09/Mar/17
![let ((n^2 +3n+1)/(3n+10))=m, m∈Z n^2 +3n+1=m(3n+10) n^2 +3(1−m)n+(1−10m)=0 n=((3(m−1)±(√(9(1−m)^2 −4(1−10m))))/2) n=((3(m−1)±(√(9−18m+9m^2 −4+40m)))/2) n=((3(m−1)±(√(9m^2 +22m+5)))/2) n=((3(m−1)±3(√(m^2 +((22)/9)m+(5/9))))/2) n=((3[m−1±(√(m^2 +((22)/9)m+(5/9)))])/2) ...(i) for n to be integer it must be m^2 +((22)/9)m+(5/9)=k^2 , k∈N m^2 +((22)/9)m+(5/9)−k^2 =0 m=((−((22)/9)±(√(((22^2 )/(81))−4×((5−9k^2 )/9))))/2) m=((−22±(√(22^2 −4×9×(5−9k^2 ))))/(2×9)) m=((−11±(√(76+(9k)^2 )))/9) ...(ii) for m to be integer it must be 76+(9k)^2 =i^2 , i∈N or (i+9k)(i−9k)=76=1×76=2×38=4×19 i−9k=1 i+9k=76 ⇒no integer solution i−9k=2 i+9k=38 ⇒i=20 and k=2 i−9k=4 i+9k=19 ⇒no integer solution ⇒the only one solution is i=20 and k=2 from (ii): ⇒m=1 or ⇒m=−((31)/9) (no integer) from (i) with m=1: ⇒n=±3](https://www.tinkutara.com/question/Q11059.png)
$${let}\:\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}{n}+\mathrm{10}}={m},\:{m}\in\mathbb{Z} \\ $$$${n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}={m}\left(\mathrm{3}{n}+\mathrm{10}\right) \\ $$$${n}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}−{m}\right){n}+\left(\mathrm{1}−\mathrm{10}{m}\right)=\mathrm{0} \\ $$$${n}=\frac{\mathrm{3}\left({m}−\mathrm{1}\right)\pm\sqrt{\mathrm{9}\left(\mathrm{1}−{m}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−\mathrm{10}{m}\right)}}{\mathrm{2}} \\ $$$${n}=\frac{\mathrm{3}\left({m}−\mathrm{1}\right)\pm\sqrt{\mathrm{9}−\mathrm{18}{m}+\mathrm{9}{m}^{\mathrm{2}} −\mathrm{4}+\mathrm{40}{m}}}{\mathrm{2}} \\ $$$${n}=\frac{\mathrm{3}\left({m}−\mathrm{1}\right)\pm\sqrt{\mathrm{9}{m}^{\mathrm{2}} +\mathrm{22}{m}+\mathrm{5}}}{\mathrm{2}} \\ $$$${n}=\frac{\mathrm{3}\left({m}−\mathrm{1}\right)\pm\mathrm{3}\sqrt{{m}^{\mathrm{2}} +\frac{\mathrm{22}}{\mathrm{9}}{m}+\frac{\mathrm{5}}{\mathrm{9}}}}{\mathrm{2}} \\ $$$${n}=\frac{\mathrm{3}\left[{m}−\mathrm{1}\pm\sqrt{{m}^{\mathrm{2}} +\frac{\mathrm{22}}{\mathrm{9}}{m}+\frac{\mathrm{5}}{\mathrm{9}}}\right]}{\mathrm{2}}\:\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${for}\:{n}\:{to}\:{be}\:{integer}\:{it}\:{must}\:{be} \\ $$$${m}^{\mathrm{2}} +\frac{\mathrm{22}}{\mathrm{9}}{m}+\frac{\mathrm{5}}{\mathrm{9}}={k}^{\mathrm{2}} ,\:\:{k}\in\mathbb{N} \\ $$$${m}^{\mathrm{2}} +\frac{\mathrm{22}}{\mathrm{9}}{m}+\frac{\mathrm{5}}{\mathrm{9}}−{k}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}=\frac{−\frac{\mathrm{22}}{\mathrm{9}}\pm\sqrt{\frac{\mathrm{22}^{\mathrm{2}} }{\mathrm{81}}−\mathrm{4}×\frac{\mathrm{5}−\mathrm{9}{k}^{\mathrm{2}} }{\mathrm{9}}}}{\mathrm{2}} \\ $$$${m}=\frac{−\mathrm{22}\pm\sqrt{\mathrm{22}^{\mathrm{2}} −\mathrm{4}×\mathrm{9}×\left(\mathrm{5}−\mathrm{9}{k}^{\mathrm{2}} \right)}}{\mathrm{2}×\mathrm{9}} \\ $$$${m}=\frac{−\mathrm{11}\pm\sqrt{\mathrm{76}+\left(\mathrm{9}{k}\right)^{\mathrm{2}} }}{\mathrm{9}}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${for}\:{m}\:{to}\:{be}\:{integer}\:{it}\:{must}\:{be} \\ $$$$\mathrm{76}+\left(\mathrm{9}{k}\right)^{\mathrm{2}} ={i}^{\mathrm{2}} ,\:\:{i}\in\mathbb{N} \\ $$$${or} \\ $$$$\left({i}+\mathrm{9}{k}\right)\left({i}−\mathrm{9}{k}\right)=\mathrm{76}=\mathrm{1}×\mathrm{76}=\mathrm{2}×\mathrm{38}=\mathrm{4}×\mathrm{19} \\ $$$$ \\ $$$${i}−\mathrm{9}{k}=\mathrm{1} \\ $$$${i}+\mathrm{9}{k}=\mathrm{76}\:\:\:\:\Rightarrow{no}\:{integer}\:{solution} \\ $$$$ \\ $$$${i}−\mathrm{9}{k}=\mathrm{2} \\ $$$${i}+\mathrm{9}{k}=\mathrm{38}\:\:\:\:\Rightarrow{i}=\mathrm{20}\:{and}\:{k}=\mathrm{2} \\ $$$$ \\ $$$${i}−\mathrm{9}{k}=\mathrm{4} \\ $$$${i}+\mathrm{9}{k}=\mathrm{19}\:\:\:\:\Rightarrow{no}\:{integer}\:{solution} \\ $$$$ \\ $$$$\Rightarrow{the}\:{only}\:{one}\:{solution}\:{is}\: \\ $$$${i}=\mathrm{20}\:{and}\:{k}=\mathrm{2} \\ $$$$ \\ $$$${from}\:\left({ii}\right): \\ $$$$\Rightarrow{m}=\mathrm{1}\:{or} \\ $$$$\Rightarrow{m}=−\frac{\mathrm{31}}{\mathrm{9}}\:\left({no}\:{integer}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{with}\:{m}=\mathrm{1}: \\ $$$$\Rightarrow{n}=\pm\mathrm{3} \\ $$
Commented by Joel576 last updated on 10/Mar/17
$${thank}\:{you}\:{very}\:{much} \\ $$