If-n-2-3n-1-is-divisible-by-3n-10-Find-out-all-possible-solution-for-n-

Question Number 11049 by Joel576 last updated on 09/Mar/17

If n^{2} + 3 n + 1 is divisible by 3 n + 10

Find out all possible solution for n

Commented by ajfour last updated on 09/Mar/17

c a n t f i g u r e a n y o t h e r t h a n \pm 3 .

Answered by mrW1 last updated on 09/Mar/17

l e t \frac{n^{2} + 3 n + 1}{3 n + 10} = m, m \in Z

n^{2} + 3 n + 1 = m (3 n + 10)

n^{2} + 3 (1 - m) n + (1 - 10 m) = 0

n = \frac{3 (m - 1) \pm \sqrt{9 {(1 - m)}^{2} - 4 (1 - 10 m)}}{2}

n = \frac{3 (m - 1) \pm \sqrt{9 - 18 m + 9 m^{2} - 4 + 40 m}}{2}

n = \frac{3 (m - 1) \pm \sqrt{9 m^{2} + 22 m + 5}}{2}

n = \frac{3 (m - 1) \pm 3 \sqrt{m^{2} + \frac{22}{9} m + \frac{5}{9}}}{2}

n = \frac{3 [m - 1 \pm \sqrt{m^{2} + \frac{22}{9} m + \frac{5}{9}}]}{2} \dots (i)

f o r n t o b e i n t e g e r i t m u s t b e

m^{2} + \frac{22}{9} m + \frac{5}{9} = k^{2}, k \in N

m^{2} + \frac{22}{9} m + \frac{5}{9} - k^{2} = 0

m = \frac{- \frac{22}{9} \pm \sqrt{\frac{22^{2}}{81} - 4 \times \frac{5 - 9 k^{2}}{9}}}{2}

m = \frac{- 22 \pm \sqrt{22^{2} - 4 \times 9 \times (5 - 9 k^{2})}}{2 \times 9}

m = \frac{- 11 \pm \sqrt{76 + {(9 k)}^{2}}}{9} \dots (i i)

f o r m t o b e i n t e g e r i t m u s t b e

76 + {(9 k)}^{2} = i^{2}, i \in N

o r

(i + 9 k) (i - 9 k) = 76 = 1 \times 76 = 2 \times 38 = 4 \times 19

i - 9 k = 1

i + 9 k = 76 \Rightarrow n o i n t e g e r s o l u t i o n

i - 9 k = 2

i + 9 k = 38 \Rightarrow i = 20 a n d k = 2

i - 9 k = 4

i + 9 k = 19 \Rightarrow n o i n t e g e r s o l u t i o n

\Rightarrow t h e o n l y o n e s o l u t i o n i s

i = 20 a n d k = 2

f r o m (i i) :

\Rightarrow m = 1 o r

\Rightarrow m = - \frac{31}{9} (n o i n t e g e r)

f r o m (i) w i t h m = 1 :

\Rightarrow n = \pm 3

Commented by Joel576 last updated on 10/Mar/17

t h a n k y o u v e r y m u c h

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