4k-1-k-3-4k-1-k-3-11-k-3-1or11-I-can-t-understand-Who-can-help-me-

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}, \dots

Commented by MJS last updated on 29/Apr/18

n = 13 seems to be the only solution \dots

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 \dots 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 \dots 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 .