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Question Number 100951 by bemath last updated on 29/Jun/20

Commented by Rasheed.Sindhi last updated on 29/Jun/20

By john santu(answer of q#97746) Same as above. 2^x (1+2^(x+1) ) = (y−1)(y+1) show that the factors y−1 and y+1 are even, exactly one of them divisible by 4. Hence x≥3 and one of these factors is divisible by 2^(x−1) but not by 2^x . so y = 2^(x−1) m+ε , m odd , ε = ±1 plugging this into the original equation we obtain 2^x (1+2^(x+1) )=(2^(x−1) m+ε)^2 −1 = 2^(2x−2) m^2 +2^x mε or equivalently 1+2^(x+1) =2^(x−2) m^2 +mε ⇔ thus we have the complete list solutions (x,y) : (0,2),(0,−2), (4,23),(4,−23).

$${By}\:{john}\:{santu}\left({answer}\:{of}\right. \\ $$$$\left.{q}#\mathrm{97746}\right)\:{Same}\:{as}\:{above}. \\ $$$$\mathrm{2}^{{x}} \left(\mathrm{1}+\mathrm{2}^{{x}+\mathrm{1}} \right)\:=\:\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right) \\ $$$${show}\:{that}\:{the}\:{factors}\:{y}−\mathrm{1}\:{and}\:{y}+\mathrm{1} \\ $$$${are}\:{even},\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\: \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\mathrm{4}.\:\mathrm{Hence}\:\mathrm{x}\geqslant\mathrm{3}\:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{factors}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{by}\:\mathrm{2}^{\mathrm{x}} .\:\mathrm{so}\:\mathrm{y}\:=\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} \mathrm{m}+\epsilon\:,\:\mathrm{m}\:\mathrm{odd}\:,\:\epsilon\:=\:\pm\mathrm{1} \\ $$$$\mathrm{plugging}\:\mathrm{this}\:\mathrm{into}\:\mathrm{the}\:\mathrm{original}\: \\ $$$$\mathrm{equation}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{2}^{\mathrm{x}} \left(\mathrm{1}+\mathrm{2}^{\mathrm{x}+\mathrm{1}} \right)=\left(\mathrm{2}^{\mathrm{x}−\mathrm{1}} \mathrm{m}+\epsilon\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$=\:\mathrm{2}^{\mathrm{2x}−\mathrm{2}} \mathrm{m}^{\mathrm{2}} +\mathrm{2}^{\mathrm{x}} \mathrm{m}\epsilon \\ $$$$\mathrm{or}\:\mathrm{equivalently}\:\mathrm{1}+\mathrm{2}^{\mathrm{x}+\mathrm{1}} =\mathrm{2}^{\mathrm{x}−\mathrm{2}} \mathrm{m}^{\mathrm{2}} +\mathrm{m}\epsilon \\ $$$$\Leftrightarrow\:\mathrm{thus}\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{complete} \\ $$$$\mathrm{list}\:\mathrm{solutions}\:\left(\mathrm{x},\mathrm{y}\right)\::\:\left(\mathrm{0},\mathrm{2}\right),\left(\mathrm{0},−\mathrm{2}\right), \\ $$$$\left(\mathrm{4},\mathrm{23}\right),\left(\mathrm{4},−\mathrm{23}\right).\: \\ $$

Commented by Rasheed.Sindhi last updated on 30/Jun/20

Sir bemath, Your solution (which you have given as a question) is solution of your q#97746 which was answered there by sirjohn santu and your this solution is same as his. Why you′ve repeated it and gave answer without question!!! (Only to satisfy my curosity) :)

$${Sir}\:{bemath}, \\ $$$${Your}\:{solution}\:\left({which}\:{you}\:{have}\right. \\ $$$$\left.{given}\:{as}\:{a}\:{question}\right)\:{is}\:{solution} \\ $$$${of}\:{your}\:{q}#\mathrm{97746}\:{which}\:{was} \\ $$$${answered}\:{there}\:{by}\:\:{sirjohn}\:{santu} \\ $$$${and}\:{your}\:{this}\:{solution}\:{is}\:{same} \\ $$$${as}\:{his}.\:{Why}\:{you}'{ve}\:{repeated}\:{it} \\ $$$${and}\:{gave}\:{answer}\:{without}\:{question}!!! \\ $$$$\left.\left({Only}\:{to}\:{satisfy}\:{my}\:{curosity}\right)\::\right) \\ $$

Commented by 1549442205 last updated on 01/Jul/20

I guess perhaps he want everyone know above his question with a good solution for it

$$\mathrm{I}\:\mathrm{guess}\:\mathrm{perhaps}\:\mathrm{he}\:\mathrm{want}\:\mathrm{everyone}\:\mathrm{know}\:\mathrm{above} \\ $$$$\mathrm{his}\:\mathrm{question}\:\mathrm{with}\:\mathrm{a}\:\:\mathrm{good}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{it} \\ $$

Commented by Rasheed.Sindhi last updated on 01/Jul/20

May be.

$$\mathcal{M}{ay}\:{be}. \\ $$

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