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Question Number 111035 by ZiYangLee last updated on 01/Sep/20

If b∈Z^+ ∀ both the roots of equation x^2 −bx+132=0 are integers. Find (1)the largest possible value of b (2)the smallest possible value of b.

$$\mathrm{If}\:{b}\in\mathbb{Z}^{+} \:\forall\:\mathrm{both}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} −{bx}+\mathrm{132}=\mathrm{0}\:\mathrm{are}\:\mathrm{integers}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{1}\right)\mathrm{the}\:\mathrm{largest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{b} \\ $$$$\left(\mathrm{2}\right)\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{b}. \\ $$

Answered by 1549442205PVT last updated on 02/Sep/20

x^2 −bx+132=(x−m)(x−n) ⇔ { ((m+n=b>0)),((mn=132=3.11.4)) :} WLOG suppose m≤n ⇒m∈{1,2,3,4,6,11} i)m=1⇒n=132⇒m+n=133 ii)m=2⇒n=66⇒m+n=68 iii)m=3⇒n=44⇒m+n=47 iv)m=4⇒n=33⇒m+n=37 v)m=6⇒n=22⇒m+n=28 vi)m=11⇒n=12⇒m+n=23 Thus,b=m+n get smallest value equal to 23 so that the given eq. x^2 −bx+132 has both its roots are integers

$$\mathrm{x}^{\mathrm{2}} −\mathrm{bx}+\mathrm{132}=\left(\mathrm{x}−\mathrm{m}\right)\left(\mathrm{x}−\mathrm{n}\right) \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{m}+\mathrm{n}=\mathrm{b}>\mathrm{0}}\\{\mathrm{mn}=\mathrm{132}=\mathrm{3}.\mathrm{11}.\mathrm{4}}\end{cases} \\ $$$$\mathrm{WLOG}\:\mathrm{suppose}\:\mathrm{m}\leqslant\mathrm{n} \\ $$$$\Rightarrow\mathrm{m}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{11}\right\} \\ $$$$\left.\mathrm{i}\right)\mathrm{m}=\mathrm{1}\Rightarrow\mathrm{n}=\mathrm{132}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{133} \\ $$$$\left.\mathrm{ii}\right)\mathrm{m}=\mathrm{2}\Rightarrow\mathrm{n}=\mathrm{66}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{68} \\ $$$$\left.\mathrm{iii}\right)\mathrm{m}=\mathrm{3}\Rightarrow\mathrm{n}=\mathrm{44}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{47} \\ $$$$\left.\mathrm{iv}\right)\mathrm{m}=\mathrm{4}\Rightarrow\mathrm{n}=\mathrm{33}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{37} \\ $$$$\left.\mathrm{v}\right)\mathrm{m}=\mathrm{6}\Rightarrow\mathrm{n}=\mathrm{22}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{28} \\ $$$$\left.\mathrm{vi}\right)\mathrm{m}=\mathrm{11}\Rightarrow\mathrm{n}=\mathrm{12}\Rightarrow\mathrm{m}+\mathrm{n}=\mathrm{23} \\ $$$$\mathrm{Thus},\mathrm{b}=\mathrm{m}+\mathrm{n}\:\mathrm{get}\:\mathrm{smallest}\:\mathrm{value} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{23}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{given}\:\mathrm{eq}. \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{bx}+\mathrm{132}\:\mathrm{has}\:\mathrm{both}\:\mathrm{its}\:\mathrm{roots}\:\mathrm{are} \\ $$$$\mathrm{integers} \\ $$

Commented by 1549442205PVT last updated on 02/Sep/20

Since m+n=b>0,but mn=132 ,so both are positive,Sir!

$$\mathrm{Since}\:\mathrm{m}+\mathrm{n}=\mathrm{b}>\mathrm{0},\mathrm{but}\:\mathrm{mn}=\mathrm{132}\:,\mathrm{so} \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{positive},\mathrm{Sir}! \\ $$

Commented by Her_Majesty last updated on 01/Sep/20

m>0∧n>0 or m<0∧n<0 ⇒ check your answer. the method is ok

$${m}>\mathrm{0}\wedge{n}>\mathrm{0}\:{or}\:{m}<\mathrm{0}\wedge{n}<\mathrm{0}\:\Rightarrow\:{check}\:{your} \\ $$$${answer}.\:{the}\:{method}\:{is}\:{ok} \\ $$

Commented by Her_Majesty last updated on 02/Sep/20

sorry I missed that b∈Z^+ , you are right

$${sorry}\:{I}\:{missed}\:{that}\:{b}\in\mathbb{Z}^{+} ,\:{you}\:{are}\:{right} \\ $$

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