((4k+1)/(k+3)),(4k+1,k+3)=(11,k+3)=1or11; I can′t understand.Who can help me?


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Question Number 34005 by math2018 last updated on 29/Apr/18

$\frac{4 k + 1}{k + 3}, (4 k + 1, k + 3) = (11, k + 3) = 1 o r 11;$ $I {c a n}^{'} t u n d e r s t a n d . W h o c a n h e l p m e ?$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18

$p l s c l a r i f y t h e q u e s t i o n o r u p l o a d t h e i m a g e o f q u e s t i o n$
Commented by math2018 last updated on 29/Apr/18

$T h e o r i g i n a l q u e s t i o n i s -$ $‘ ‘ I f \frac{4 n - 2}{n + 5} i s t h e s q u a r e o f a r a t i o n a l n u m b e r,$ $f i n d a l l i n t e g e r s n p l s .$ $A n s w e r :$ $(1) \frac{4 n - 2}{n + 5} = \frac{2 (2 n - 1)}{n + 5};$ $(2) l e t n = 2 k + 1, t h e n w e h a v e$ $\frac{4 n - 2}{n + 5} = \frac{4 k + 1}{k + 3}, . . .$
Commented by MJS last updated on 29/Apr/18
$n=13 seems to be the only solution... two possibilties 1. ((4k+1)/(k+3)) is a reduced fraction ⇒ 4k+1 and k+3 both must be square integers 4k+1=p^2 k+3=q^2 ⇒ k=q^2 −3 4(q^2 −3)+1=p^2 ⇒ p^2 =4q^2 −11 ⇒ p^2 =(2q)^2 −11=r^2 −11 so we′re looking for 2 integers with r^2 −p^2 =11 ⇒ r=6 ∧ p=5 ⇒ q=3 ⇒ ⇒ k=6 ⇒ n=13 2. ((4k+1)/(k+3)) can be reduced by factor f≠±1 4k+1=f×u ⇒ k=((fu−1)/4) (I.) k+3=f×v ⇒ k=fv−3 (II.) ((fu−1)/4)=fv−3 ⇒ f=((11)/(4v−u)); f, u, v ∈Z\{0} ⇒ ⇒ 4v−u=±11 2.1. 4v−u=−11 ⇒ u=4v+11 (I.)=(II.) ((−11(4v+11)−1)/4)=−11v−3 −11v−((61)/2)=−11v−3 ⇒ no solution 2.2. 4v−u=11 ⇒ u=4v−11 (I.)=(II.) ((11(4v−11)−1)/4)=11v−3 11v−((61)/2)=11v−3 ⇒ no solution$
$n = 13 seems to be the only solution . . .$ $two possibilties$ $1 . \frac{4 k + 1}{k + 3} is a reduced fraction$ $\Rightarrow 4 k + 1 and k + 3 both must be square integers$ $4 k + 1 = p^{2}$ $k + 3 = q^{2} \Rightarrow k = q^{2} - 3$ $4 (q^{2} - 3) + 1 = p^{2} \Rightarrow p^{2} = 4 q^{2} - 11$ $\Rightarrow p^{2} = {(2 q)}^{2} - 11 = r^{2} - 11$ $so {we}^{'} re looking for 2 integers with$ $r^{2} - p^{2} = 11 \Rightarrow r = 6 \land p = 5 \Rightarrow q = 3 \Rightarrow$ $\Rightarrow k = 6 \Rightarrow n = 13$ $2 . \frac{4 k + 1}{k + 3} can be reduced by factor f \neq \pm 1$ $4 k + 1 = f \times u \Rightarrow k = \frac{f u - 1}{4} (I .)$ $k + 3 = f \times v \Rightarrow k = f v - 3 (I I .)$ $\frac{f u - 1}{4} = f v - 3 \Rightarrow f = \frac{11}{4 v - u}; f, u, v \in Z ∖ {0} \Rightarrow$ $\Rightarrow 4 v - u = \pm 11$ $2.1 .$ $4 v - u = - 11 \Rightarrow u = 4 v + 11$ $(I .) = (I I .)$ $\frac{- 11 (4 v + 11) - 1}{4} = - 11 v - 3$ $- 11 v - \frac{61}{2} = - 11 v - 3 \Rightarrow no solution$ $2.2 .$ $4 v - u = 11 \Rightarrow u = 4 v - 11$ $(I .) = (I I .)$ $\frac{11 (4 v - 11) - 1}{4} = 11 v - 3$ $11 v - \frac{61}{2} = 11 v - 3 \Rightarrow no solution$
Commented by math2018 last updated on 29/Apr/18

$T h a n k s S i r!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18

	$a s p e r y o u r c o m m e n t \frac{4 n - 2}{n + 5} i s t h e s q u a r e d$ $v a l u e o f a r a t i o n a l n u m b e r$ $l e t u s e x a m i n e$ $u n i t p l a c e o f a n u m b e r u n i t p l a c e o f s q u a r e d$ $1 1$ $2 4$ $3 9$ $4 6$ $5 5$ $6 6$ $7 9$ $8 4$ $9 1$ $4 n - 2 i s e v e n s o u n i t p l a c e o f N_{r} e i t h e r 4 o r 6$ $n + 5 o d d s o u n i t p l a c e m a y b e 1 o r 5 o r 9$ $l e t m e t h i n k . . . i s t h e r e a n y o t h e r i n f o r m a t i o n i n [t$ $i n t h e q u e s t i o n . . . p l s c h e c k$
	Commented by math2018 last updated on 30/Apr/18

	$T h a n k y o u S i r . T h e r i g h t a n s w e r i s n = 13 a n d t h e r e i s n o m o r e i n f o r m a t i o n t o a d d .$
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