Question Number 217030 by ArshadS last updated on 27/Feb/25
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that}\:\: \\ $$$$\:\mathrm{n}\:+\:\mathrm{1}\:\:\mathrm{divides}\:\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$
Answered by issac last updated on 27/Feb/25
$$\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\frac{{n}+\mathrm{1}}{\left(\mathrm{1}+{n}\boldsymbol{{i}}\right)\left(\mathrm{1}−{n}\boldsymbol{{i}}\right)} \\ $$$$\frac{{n}+\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+{n}\boldsymbol{{i}}}+\frac{\mathrm{1}}{\mathrm{1}−{n}\boldsymbol{{i}}}\right) \\ $$$${n}=\mathrm{2}\mathbb{Z}−\mathrm{1}\: \\ $$
Commented by Frix last updated on 27/Feb/25
$$“{a}\:\mathrm{divides}\:{b}''\:\mathrm{means}\:{a}×{n}={b}\:\:\Leftrightarrow\:\frac{{b}}{{a}}={n}\:\mathrm{not} \\ $$$$\cancel{{b}×{n}={a}\:\Leftrightarrow\:\frac{{a}}{{b}}={n}} \\ $$$$\mathrm{You}\:\mathrm{must}\:\mathrm{test}\:\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{1}}\:\mathrm{not}\:\cancel{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Answered by Ghisom last updated on 27/Feb/25
$$\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{1}}={n}−\mathrm{1}+\frac{\mathrm{2}}{{n}+\mathrm{1}} \\ $$$$\mathrm{for}\:{n}\in\mathbb{Z}\:\mathrm{we}\:\mathrm{get}\:{n}\in\left\{−\mathrm{3},\:−\mathrm{2},\:\mathrm{0},\:\mathrm{1}\right\} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{for}\:{n}>\mathrm{0}\:\mathrm{is}\:{n}=\mathrm{1} \\ $$
Commented by ArshadS last updated on 27/Feb/25
$${nice}!\:{thank}\:{you}\:{sir}! \\ $$