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Question Number 112096 by mathdave last updated on 06/Sep/20

Answered by mr W last updated on 06/Sep/20

say ab=k^2 ((a+b)/2)−k=1 ⇒a+b=2(k+1) a, b are roots of z^2 −2(k+1)z+k^2 =0 a, b=(k+1)±(√(2k+1)) 2k+1=m^2 2k=(m−1)(m+1) m must be odd: m=2n+1 2k=2n(2n+2) ⇒k=2n(n+1) a,b=2n(n+1)+1±(2n+1) say, a<b, a=2n(n+1)+1−(2n+1)=2n^2 b=2n(n+1)+1+(2n+1)=2(n+1)^2 general solution: { ((a=2n^2 )),((b=2(n+1)^2 )) :} with n∈N

$${say}\:{ab}={k}^{\mathrm{2}} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}−{k}=\mathrm{1} \\ $$$$\Rightarrow{a}+{b}=\mathrm{2}\left({k}+\mathrm{1}\right) \\ $$$${a},\:{b}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}\left({k}+\mathrm{1}\right){z}+{k}^{\mathrm{2}} =\mathrm{0} \\ $$$${a},\:{b}=\left({k}+\mathrm{1}\right)\pm\sqrt{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{2}{k}+\mathrm{1}={m}^{\mathrm{2}} \\ $$$$\mathrm{2}{k}=\left({m}−\mathrm{1}\right)\left({m}+\mathrm{1}\right) \\ $$$${m}\:{must}\:{be}\:{odd}:\:{m}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{2}{k}=\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{2}\right) \\ $$$$\Rightarrow{k}=\mathrm{2}{n}\left({n}+\mathrm{1}\right) \\ $$$${a},{b}=\mathrm{2}{n}\left({n}+\mathrm{1}\right)+\mathrm{1}\pm\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${say},\:{a}<{b}, \\ $$$${a}=\mathrm{2}{n}\left({n}+\mathrm{1}\right)+\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{2}{n}^{\mathrm{2}} \\ $$$${b}=\mathrm{2}{n}\left({n}+\mathrm{1}\right)+\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${general}\:{solution}: \\ $$$$\begin{cases}{{a}=\mathrm{2}{n}^{\mathrm{2}} }\\{{b}=\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\end{cases}\:\:{with}\:{n}\in\mathbb{N} \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by 1549442205PVT last updated on 06/Sep/20

((a+b)/2)−(√(ab)) =1⇔a+b−2=2(√(ab)) a^2 +b^2 +4+2ab−4a−4b=4ab a^2 −2(b+2)a+b^2 −4b+4=0 (1) We look at this is a quadratic equation with respect to unknown a.Then Δ′=(b+2)^2 −(b−2)^2 =8b. The equation (1) has integer roots ,so (√(Δ′))=m^2 ⇒8b=m^2 ⇒b=2u^2 .Replace into (1) we have a=2u^2 +2±4u.We have 2u^2 +4u+2>0 and 2u^2 −4u+2=2(u−1)^2 ≥0 ∀u∈N Thus,the solutions of our problem are: (a,b)={(2u^2 ±4u+2,2u^2 ),(2v^2 ,2v^2 ±4v+2)} where u,v∈N

$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\sqrt{\mathrm{ab}}\:=\mathrm{1}\Leftrightarrow\mathrm{a}+\mathrm{b}−\mathrm{2}=\mathrm{2}\sqrt{\mathrm{ab}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{4}+\mathrm{2ab}−\mathrm{4a}−\mathrm{4b}=\mathrm{4ab} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{b}+\mathrm{2}\right)\mathrm{a}+\mathrm{b}^{\mathrm{2}} −\mathrm{4b}+\mathrm{4}=\mathrm{0}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{We}\:\mathrm{look}\:\mathrm{at}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{unknown}\:\mathrm{a}.\mathrm{Then} \\ $$$$\Delta'=\left(\mathrm{b}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{b}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{8b}. \\ $$$$\mathrm{The}\:\mathrm{equation}\:\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{integer}\:\mathrm{roots}\:,\mathrm{so} \\ $$$$\sqrt{\Delta'}=\mathrm{m}^{\mathrm{2}} \Rightarrow\mathrm{8b}=\mathrm{m}^{\mathrm{2}} \Rightarrow\mathrm{b}=\mathrm{2u}^{\mathrm{2}} .\mathrm{Replace} \\ $$$$\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{a}=\mathrm{2u}^{\mathrm{2}} +\mathrm{2}\pm\mathrm{4u}.\mathrm{We}\:\mathrm{have}\:\mathrm{2u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{2}>\mathrm{0} \\ $$$$\:\mathrm{and}\:\mathrm{2u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{2}=\mathrm{2}\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\forall\mathrm{u}\in\mathbb{N} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{our}\:\mathrm{problem}\:\mathrm{are}: \\ $$$$\left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\right)=\left\{\left(\mathrm{2}\boldsymbol{\mathrm{u}}^{\mathrm{2}} \pm\mathrm{4}\boldsymbol{\mathrm{u}}+\mathrm{2},\mathrm{2}\boldsymbol{\mathrm{u}}^{\mathrm{2}} \right),\left(\mathrm{2}\boldsymbol{\mathrm{v}}^{\mathrm{2}} ,\mathrm{2}\boldsymbol{\mathrm{v}}^{\mathrm{2}} \pm\mathrm{4}\boldsymbol{\mathrm{v}}+\mathrm{2}\right)\right\} \\ $$$$\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{u}},\boldsymbol{\mathrm{v}}\in\mathbb{N} \\ $$

Commented by mathdave last updated on 06/Sep/20

good solution

$${good}\:{solution} \\ $$

Commented by 1549442205PVT last updated on 06/Sep/20

Thank Sir.You are welcome.

$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

Answered by MJS_new last updated on 06/Sep/20

same solution but different path ((a+b)/2)−(√(ab))=1 a=u−v∧b=u+v with u, v ∈N ∧ u>v u−(√(u^2 −v^2 ))=1 u=((v^2 +1)/2) a=(((v−1)^2 )/2)∧b=(((v+1)^2 )/2) a, b ∈N ⇒ v=2n+1 a=2n^2 ∧b=2(n+1)^2 solution is { (((2n^2 )),((2(n+1)^2 )) ) ∣ n∈N}

$$\mathrm{same}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{different}\:\mathrm{path} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}−\sqrt{{ab}}=\mathrm{1} \\ $$$${a}={u}−{v}\wedge{b}={u}+{v}\:{with}\:{u},\:{v}\:\in\mathbb{N}\:\wedge\:{u}>{v} \\ $$$${u}−\sqrt{{u}^{\mathrm{2}} −{v}^{\mathrm{2}} }=\mathrm{1} \\ $$$${u}=\frac{{v}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}} \\ $$$${a}=\frac{\left({v}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}\wedge{b}=\frac{\left({v}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${a},\:{b}\:\in\mathbb{N}\:\Rightarrow\:{v}=\mathrm{2}{n}+\mathrm{1} \\ $$$${a}=\mathrm{2}{n}^{\mathrm{2}} \wedge{b}=\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{solution}\:\mathrm{is}\:\left\{\begin{pmatrix}{\mathrm{2}{n}^{\mathrm{2}} }\\{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\end{pmatrix}\:\mid\:{n}\in\mathbb{N}\right\} \\ $$

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