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Question Number 34005 by math2018 last updated on 29/Apr/18
((4k+1)/(k+3)),(4k+1,k+3)=(11,k+3)=1or11; I can′t understand.Who can help me?
$$\frac{\mathrm{4}{k}+\mathrm{1}}{{k}+\mathrm{3}},\left(\mathrm{4}{k}+\mathrm{1},{k}+\mathrm{3}\right)=\left(\mathrm{11},{k}+\mathrm{3}\right)=\mathrm{1}{or}\mathrm{11}; \\ $$$${I}\:{can}'{t}\:{understand}.{Who}\:{can}\:{help}\:{me}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18
pls clarify the question or upload the image of question
$${pls}\:{clarify}\:{the}\:{question}\:{or}\:{upload}\:{the}\:{image}\:{of}\:{question} \\ $$
Commented by math2018 last updated on 29/Apr/18
The original question is − “If ((4n−2)/(n+5)) is the square of a rational number, find all integers n pls. Answer: (1)((4n−2)/(n+5))=((2(2n−1))/(n+5)); (2)let n=2k+1,then we have ((4n−2)/(n+5))=((4k+1)/(k+3)),...
$${The}\:{original}\:{question}\:{is}\:− \\ $$$$“{If}\:\:\frac{\mathrm{4}{n}−\mathrm{2}}{{n}+\mathrm{5}}\:{is}\:{the}\:{square}\:{of}\:{a}\:{rational}\:{number}, \\ $$$${find}\:{all}\:{integers}\:{n}\:{pls}. \\ $$$${Answer}: \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{4}{n}−\mathrm{2}}{{n}+\mathrm{5}}=\frac{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}{{n}+\mathrm{5}}; \\ $$$$\left(\mathrm{2}\right){let}\:{n}=\mathrm{2}{k}+\mathrm{1},{then}\:{we}\:{have}\: \\ $$$$\frac{\mathrm{4}{n}−\mathrm{2}}{{n}+\mathrm{5}}=\frac{\mathrm{4}{k}+\mathrm{1}}{{k}+\mathrm{3}},… \\ $$$$ \\ $$
Commented by MJS last updated on 29/Apr/18
n=13 seems to be the only solution... two possibilties 1. ((4k+1)/(k+3)) is a reduced fraction ⇒ 4k+1 and k+3 both must be square integers 4k+1=p^2 k+3=q^2 ⇒ k=q^2 −3 4(q^2 −3)+1=p^2 ⇒ p^2 =4q^2 −11 ⇒ p^2 =(2q)^2 −11=r^2 −11 so we′re looking for 2 integers with r^2 −p^2 =11 ⇒ r=6 ∧ p=5 ⇒ q=3 ⇒ ⇒ k=6 ⇒ n=13 2. ((4k+1)/(k+3)) can be reduced by factor f≠±1 4k+1=f×u ⇒ k=((fu−1)/4) (I.) k+3=f×v ⇒ k=fv−3 (II.) ((fu−1)/4)=fv−3 ⇒ f=((11)/(4v−u)); f, u, v ∈Z\{0} ⇒ ⇒ 4v−u=±11 2.1. 4v−u=−11 ⇒ u=4v+11 (I.)=(II.) ((−11(4v+11)−1)/4)=−11v−3 −11v−((61)/2)=−11v−3 ⇒ no solution 2.2. 4v−u=11 ⇒ u=4v−11 (I.)=(II.) ((11(4v−11)−1)/4)=11v−3 11v−((61)/2)=11v−3 ⇒ no solution
$${n}=\mathrm{13}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}… \\ $$$$ \\ $$$$\mathrm{two}\:\mathrm{possibilties} \\ $$$$\mathrm{1}.\:\frac{\mathrm{4}{k}+\mathrm{1}}{{k}+\mathrm{3}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{reduced}\:\mathrm{fraction} \\ $$$$\Rightarrow\:\mathrm{4}{k}+\mathrm{1}\:\mathrm{and}\:\mathrm{k}+\mathrm{3}\:\mathrm{both}\:\mathrm{must}\:\mathrm{be}\:\mathrm{square}\:\mathrm{integers} \\ $$$$\mathrm{4}{k}+\mathrm{1}={p}^{\mathrm{2}} \\ $$$${k}+\mathrm{3}={q}^{\mathrm{2}} \:\Rightarrow\:{k}={q}^{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{4}\left({q}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{1}={p}^{\mathrm{2}} \:\Rightarrow\:{p}^{\mathrm{2}} =\mathrm{4}{q}^{\mathrm{2}} −\mathrm{11} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} =\left(\mathrm{2}{q}\right)^{\mathrm{2}} −\mathrm{11}={r}^{\mathrm{2}} −\mathrm{11} \\ $$$$\mathrm{so}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{2}\:\mathrm{integers}\:\mathrm{with} \\ $$$${r}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{11}\:\Rightarrow\:{r}=\mathrm{6}\:\wedge\:{p}=\mathrm{5}\:\Rightarrow\:{q}=\mathrm{3}\:\Rightarrow \\ $$$$\Rightarrow\:{k}=\mathrm{6}\:\Rightarrow\:{n}=\mathrm{13} \\ $$$$ \\ $$$$\mathrm{2}.\:\frac{\mathrm{4}{k}+\mathrm{1}}{{k}+\mathrm{3}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{reduced}\:\mathrm{by}\:\mathrm{factor}\:{f}\neq\pm\mathrm{1} \\ $$$$\mathrm{4}{k}+\mathrm{1}={f}×{u}\:\Rightarrow\:{k}=\frac{{fu}−\mathrm{1}}{\mathrm{4}}\:\left({I}.\right) \\ $$$${k}+\mathrm{3}={f}×{v}\:\Rightarrow\:{k}={fv}−\mathrm{3}\:\left({II}.\right) \\ $$$$\frac{{fu}−\mathrm{1}}{\mathrm{4}}={fv}−\mathrm{3}\:\Rightarrow\:{f}=\frac{\mathrm{11}}{\mathrm{4}{v}−{u}};\:{f},\:{u},\:{v}\:\in\mathbb{Z}\backslash\left\{\mathrm{0}\right\}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{4}{v}−{u}=\pm\mathrm{11} \\ $$$$\mathrm{2}.\mathrm{1}. \\ $$$$\mathrm{4}{v}−{u}=−\mathrm{11}\:\Rightarrow\:{u}=\mathrm{4}{v}+\mathrm{11} \\ $$$$\left({I}.\right)=\left({II}.\right) \\ $$$$\frac{−\mathrm{11}\left(\mathrm{4}{v}+\mathrm{11}\right)−\mathrm{1}}{\mathrm{4}}=−\mathrm{11}{v}−\mathrm{3} \\ $$$$−\mathrm{11}{v}−\frac{\mathrm{61}}{\mathrm{2}}=−\mathrm{11}{v}−\mathrm{3}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{2}.\mathrm{2}. \\ $$$$\mathrm{4}{v}−{u}=\mathrm{11}\:\Rightarrow\:{u}=\mathrm{4}{v}−\mathrm{11} \\ $$$$\left({I}.\right)=\left({II}.\right) \\ $$$$\frac{\mathrm{11}\left(\mathrm{4}{v}−\mathrm{11}\right)−\mathrm{1}}{\mathrm{4}}=\mathrm{11}{v}−\mathrm{3} \\ $$$$\mathrm{11}{v}−\frac{\mathrm{61}}{\mathrm{2}}=\mathrm{11}{v}−\mathrm{3}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$
Commented by math2018 last updated on 29/Apr/18
Thanks Sir!
$${Thanks}\:{Sir}! \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18
as per your comment((4n−2)/(n+5)) is the squared value of a rational number let us examine unit place of a number unit place of squared 1 1 2 4 3 9 4 6 5 5 6 6 7 9 8 4 9 1 4n−2 is even so unit place of N_(r ) either 4 or 6 n+5 odd so unit place may be 1 or 5 or 9 let me think...is there any other information in[t in the question...pls check
$${as}\:{per}\:{your}\:{comment}\frac{\mathrm{4}{n}−\mathrm{2}}{{n}+\mathrm{5}}\:{is}\:{the}\:{squared}\: \\ $$$${value}\:{of}\:{a}\:{rational}\:{number} \\ $$$${let}\:{us}\:{examine} \\ $$$${unit}\:{place}\:{of}\:{a}\:{number}\:\:\:\:\:\:\:\:\:{unit}\:{place}\:{of}\:{squared} \\ $$$$\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9} \\ $$$$\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5} \\ $$$$\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9} \\ $$$$\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{4}{n}−\mathrm{2}\:{is}\:{even}\:{so}\:{unit}\:{place}\:{of}\:{N}_{{r}\:} {either}\:\mathrm{4}\:{or}\:\mathrm{6} \\ $$$${n}+\mathrm{5}\:{odd}\:{so}\:{unit}\:{place}\:{may}\:{be}\:\mathrm{1}\:{or}\:\mathrm{5}\:{or}\:\mathrm{9} \\ $$$${let}\:{me}\:{think}…{is}\:{there}\:{any}\:{other}\:{information}\:{in}\left[{t}\right. \\ $$$${in}\:{the}\:{question}…{pls}\:{check} \\ $$$$ \\ $$
Commented by math2018 last updated on 30/Apr/18
Thank you Sir.The right answer is n=13 and there is no more information to add.
$${Thank}\:{you}\:{Sir}.{The}\:{right}\:{answer}\:{is}\:{n}=\mathrm{13}\:{and}\:{there}\:{is}\:{no}\:{more}\:{information}\:{to}\:{add}. \\ $$

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