Let-S-n-n-2-20n-12-n-a-positive-integer-What-is-the-sum-of-all-possible-values-of-n-for-which-S-n-is-a-perfect-square-

Question Number 19332 by Tinkutara last updated on 09/Aug/17

Let S_{n} = n^{2} + 20 n + 12, n a positive

integer . What is the sum of all possible

values of n for which S_{n} is a perfect

square ?

Commented by mrW1 last updated on 09/Aug/17

Σ n = 3 + 13 = 16

Answered by mrW1 last updated on 09/Aug/17

n^{2} + 20 n + 12 = m^{2}

n^{2} + 20 n + 12 - m^{2} = 0

\Rightarrow n = \frac{- 20 \pm \sqrt{20^{2} - 4 (12 - m^{2})}}{2} = - 10 \pm \sqrt{88 + m^{2}}

88 + m^{2} = k^{2}

k^{2} - m^{2} = 88

(k - m) (k + m) = 88 = a \times b

88 = 2^{3} \times 11

a \times b = 1 \times 88 = 2 \times 44 = 4 \times 22 = 8 \times 11

k - m = a

k + m = b

\Rightarrow k = \frac{a + b}{2}

for k to be integer, a and b must be

both odd or both even .

\Rightarrow k = \frac{2 + 44}{2} = 23

\Rightarrow k = \frac{4 + 22}{2} = 13

\Rightarrow n = - 10 \pm 23 = - 33, 13

\Rightarrow n = - 10 \pm 13 = - 23, 3

sum of + ve n = 3 + 13 = 16

Commented by Tinkutara last updated on 09/Aug/17

Thank you very much Sir!