Question Number 207436 by CrispyXYZ last updated on 15/May/24
![a, b∈N_+ , ((b+1)/a)+((a+1)/b)∈Z. Prove that (a, b)≤(√(a+b.))](https://www.tinkutara.com/question/Q207436.png)
$${a},\:{b}\in\mathbb{N}_{+} ,\:\frac{{b}+\mathrm{1}}{{a}}+\frac{{a}+\mathrm{1}}{{b}}\in\mathbb{Z}.\:\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left({a},\:{b}\right)\leqslant\sqrt{{a}+{b}.} \\ $$
Commented by A5T last updated on 15/May/24
![What is meant by (a,b)?](https://www.tinkutara.com/question/Q207439.png)
$${What}\:{is}\:{meant}\:{by}\:\left({a},{b}\right)? \\ $$
Commented by CrispyXYZ last updated on 15/May/24
![the greatest common divisor of a, b](https://www.tinkutara.com/question/Q207440.png)
$$\mathrm{the}\:\mathrm{greatest}\:\mathrm{common}\:\mathrm{divisor}\:\mathrm{of}\:{a},\:{b} \\ $$
Answered by Berbere last updated on 16/May/24
![((b^2 +a^2 +a+b)/(ab))∈Z ab∣(a+b)^2 −ab+(a+b);a∣b⇔a=kb ⇒ab∣(a+b)(1+(a+b))′..(E) letd=(a,b)⇒a=da′;b=db′ E⇔d^2 a′b′∣(da′+db′)(1+d(a′+b′)) d^2 ∣d^2 a′b′⇒d^2 ∣(da′+db′)+d^2 (a′+b′)^2 ⇒d^2 ∣d(a′+b′)⇔d^2 ∣da′+db′⇔d^2 ∣a+b ⇒d^2 ≤a+b⇒d≤(√(a+b))](https://www.tinkutara.com/question/Q207467.png)
$$\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{a}+{b}}{{ab}}\in{Z} \\ $$$${ab}\mid\left({a}+{b}\right)^{\mathrm{2}} −{ab}+\left({a}+{b}\right);{a}\mid{b}\Leftrightarrow{a}={kb} \\ $$$$\Rightarrow{ab}\mid\left({a}+{b}\right)\left(\mathrm{1}+\left({a}+{b}\right)\right)'..\left({E}\right) \\ $$$${letd}=\left({a},{b}\right)\Rightarrow{a}={da}';{b}={db}' \\ $$$${E}\Leftrightarrow{d}^{\mathrm{2}} {a}'{b}'\mid\left({da}'+{db}'\right)\left(\mathrm{1}+{d}\left({a}'+{b}'\right)\right) \\ $$$${d}^{\mathrm{2}} \mid{d}^{\mathrm{2}} {a}'{b}'\Rightarrow{d}^{\mathrm{2}} \mid\left({da}'+{db}'\right)+{d}^{\mathrm{2}} \left({a}'+{b}'\right)^{\mathrm{2}} \\ $$$$\Rightarrow{d}^{\mathrm{2}} \mid{d}\left({a}'+{b}'\right)\Leftrightarrow{d}^{\mathrm{2}} \mid{da}'+{db}'\Leftrightarrow{d}^{\mathrm{2}} \mid{a}+{b} \\ $$$$\Rightarrow{d}^{\mathrm{2}} \leqslant{a}+{b}\Rightarrow{d}\leqslant\sqrt{{a}+{b}} \\ $$$$ \\ $$