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Q-168480-reposted-n-2-n-109-x-2-x-Z-n-Z-

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Question Number 168823 by Rasheed.Sindhi last updated on 18/Apr/22
Q#168480 reposted. n^2 +n+109=x^2 x∈Z, n(∈Z^+ )=?
$$\boldsymbol{\mathrm{Q}}#\mathrm{168480}\:\mathrm{reposted}. \\ $$$$\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{n}}+\mathrm{109}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}\in\mathbb{Z},\:\boldsymbol{\mathrm{n}}\left(\in\mathbb{Z}^{+} \right)=? \\ $$
Answered by Rasheed.Sindhi last updated on 18/Apr/22
n^2 +n+109=x^2 x∈Z, n(∈Z^+ )=? There′s no harm to assume x as non-negative because x^2 is anyway non-negative. n^2 +n+109−x^2 =0 n=((−1±(√(1−4(109−x^2 ))))/2) n=((−1±(√(4x^2 −435)))/2) ∵ n>0 ∴ n=((−1+(√(4x^2 −435)))/2) ∵n∈Z^+ ∴4x^2 −435 { ((>1)),(( perfect square)) :} x^2 >((436)/4)⇒x^2 ≥109⇒x≥11 determinant ((x,((√(4x^2 −435)) ),(n=((−1+(√(4x^2 −435)))/2))),((11),7,3),((23),(41),(20)),((37),(71),(35)),((109),(217),(108)),((...),(...),(...)))
$$\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{n}}+\mathrm{109}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}\in\mathbb{Z},\:\boldsymbol{\mathrm{n}}\left(\in\mathbb{Z}^{+} \right)=? \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{harm}\:\mathrm{to}\:\mathrm{assume}\:\mathrm{x}\:\mathrm{as}\: \\ $$$$\mathrm{non}-\mathrm{negative}\:\mathrm{because}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{anyway} \\ $$$$\mathrm{non}-\mathrm{negative}. \\ $$$$\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{n}}+\mathrm{109}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{n}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{109}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{n}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{435}}}{\mathrm{2}} \\ $$$$\:\because\:\boldsymbol{\mathrm{n}}>\mathrm{0} \\ $$$$\:\therefore\:\:\:\:\boldsymbol{\mathrm{n}}=\frac{−\mathrm{1}+\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{435}}}{\mathrm{2}} \\ $$$$\because\boldsymbol{\mathrm{n}}\in\mathbb{Z}^{+} \\ $$$$\therefore\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{435\begin{cases}{>\mathrm{1}}\\{\:\mathrm{perfect}\:\mathrm{square}}\end{cases}}\: \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{2}} >\frac{\mathrm{436}}{\mathrm{4}}\Rightarrow\mathrm{x}^{\mathrm{2}} \geqslant\mathrm{109}\Rightarrow\mathrm{x}\geqslant\mathrm{11} \\ $$$$\begin{array}{|c|c|c|c|c|c|}{\mathrm{x}}&\hline{\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{435}}\:}&\hline{\boldsymbol{\mathrm{n}}=\frac{−\mathrm{1}+\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{435}}}{\mathrm{2}}}\\{\mathrm{11}}&\hline{\mathrm{7}}&\hline{\mathrm{3}}\\{\mathrm{23}}&\hline{\mathrm{41}}&\hline{\mathrm{20}}\\{\mathrm{37}}&\hline{\mathrm{71}}&\hline{\mathrm{35}}\\{\mathrm{109}}&\hline{\mathrm{217}}&\hline{\mathrm{108}}\\{…}&\hline{…}&\hline{…}\\\hline\end{array} \\ $$
Answered by mindispower last updated on 18/Apr/22
⇔4n^2 +4n+436=(2x)^2 ⇔(2n+1)^2 +435=(2x)^2 ⇔(2x−2n−1)(2x+2n+1)=435 Solve for x∈Z_+ 2x+2n+1>2x−2n−1,x>n 435=5.3.29 ⇔ { ((2x+2n+1=435)),((2x−2n−1=1)) :} x=109,n=108 { ((2x+2n+1=87)),((2x−2n−1=5)) :} x=23 n=20 { ((2x+2n+1=145)),((2x−2n−1=3)) :} ⇒x=37,n=35 { ((2x+2n+1=29)),((2x−2n−1=15)) :} x=11,n=3 S=(x,n)∈{(11,3);(23,20);(37,35);(109,108);(−11,3), (−23,20);(−37,35);(−109,108)}
$$\Leftrightarrow\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{436}=\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{435}=\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{435} \\ $$$${Solve}\:{for}\:{x}\in\mathbb{Z}_{+} \\ $$$$\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}>\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1},{x}>{n} \\ $$$$\mathrm{435}=\mathrm{5}.\mathrm{3}.\mathrm{29} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}=\mathrm{435}}\\{\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1}=\mathrm{1}}\end{cases} \\ $$$${x}=\mathrm{109},{n}=\mathrm{108} \\ $$$$\begin{cases}{\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}=\mathrm{87}}\\{\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1}=\mathrm{5}}\end{cases} \\ $$$${x}=\mathrm{23} \\ $$$${n}=\mathrm{20} \\ $$$$\begin{cases}{\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}=\mathrm{145}}\\{\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1}=\mathrm{3}}\end{cases} \\ $$$$\Rightarrow{x}=\mathrm{37},{n}=\mathrm{35} \\ $$$$\begin{cases}{\mathrm{2}{x}+\mathrm{2}{n}+\mathrm{1}=\mathrm{29}}\\{\mathrm{2}{x}−\mathrm{2}{n}−\mathrm{1}=\mathrm{15}}\end{cases} \\ $$$${x}=\mathrm{11},{n}=\mathrm{3} \\ $$$${S}=\left({x},{n}\right)\in\left\{\left(\mathrm{11},\mathrm{3}\right);\left(\mathrm{23},\mathrm{20}\right);\left(\mathrm{37},\mathrm{35}\right);\left(\mathrm{109},\mathrm{108}\right);\left(−\mathrm{11},\mathrm{3}\right),\right. \\ $$$$\left.\left(−\mathrm{23},\mathrm{20}\right);\left(−\mathrm{37},\mathrm{35}\right);\left(−\mathrm{109},\mathrm{108}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 18/Apr/22
∨ ∩i⊂∈ solution!
$$\vee\:\cap\mathrm{i}\subset\in\:\mathrm{solution}! \\ $$
Commented by Rasheed.Sindhi last updated on 18/Apr/22
Thanks sir!
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by mindispower last updated on 19/Apr/22
withe Pleasur Have a nice Day
$${withe}\:{Pleasur}\:{Have}\:{a}\:{nice}\:{Day} \\ $$
Commented by Rasheed.Sindhi last updated on 19/Apr/22
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