# Let S_n = n^2 + 20n + 12, n is a positive integer. What is the sum of all the possible values of n for which S_n is a perfect square?

##### 1 Answer
Nov 29, 2016

Given

${S}_{n} = {n}^{2} + 20 n + 12 ,$
$\text{where "n = +ve " integer}$

Given expression can be arranged in different ways associated with a perfect square of integers.Here only 12 arrangements have been shown.

${S}_{n} = {\left(n + 1\right)}^{2} + 18 n + 11. \ldots \ldots . . \left[1\right]$

${S}_{n} = {\left(n + 2\right)}^{2} + 16 n + 8. \ldots \ldots \ldots \left[2\right]$

${S}_{n} = {\left(n + 3\right)}^{2} + 14 n + 3. \ldots \ldots \ldots \left[3\right]$

${S}_{n} = {\left(n + 4\right)}^{2} + 12 n - 4. \ldots \ldots \ldots \left[4\right]$

${S}_{n} = {\left(n + 5\right)}^{2} + 10 n - 13. \ldots \ldots . . \left[5\right]$

${S}_{n} = {\left(n + 6\right)}^{2} + \textcolor{red}{8 \left(n - 3\right) \ldots \ldots \ldots \left[6\right]}$

${S}_{n} = {\left(n + 7\right)}^{2} + 6 n - 37. \ldots \ldots \ldots \left[7\right]$

${S}_{n} = {\left(n + 8\right)}^{2} + \textcolor{red}{4 \left(n - 13\right) \ldots \ldots \ldots \left[8\right]}$

${S}_{n} = {\left(n + 9\right)}^{2} + 2 n - 69. \ldots \ldots \ldots \left[9\right]$

${S}_{n} = {\left(n + 10\right)}^{2} - 88. \ldots \ldots \ldots \ldots . \left[10\right]$

${S}_{n} = {\left(n + 11\right)}^{2} - 2 n - 109. \ldots \ldots . . \left[11\right]$

${S}_{n} = {\left(n + 12\right)}^{2} - 4 \left(n + 33\right) \ldots \ldots \ldots \left[12\right]$

On inspection of above 10 relations we see that ${S}_{n}$ will be perfect square in two cases i.e 6th and 8th,when n=3 and n=13 respectively.

So the sum of all the possible values of n for which ${S}_{n}$ is a perfect square is =(3+13)=16.

${S}_{n}$ may be a perfect square other than these two for negatve value of n. Case 12 where $n = - 33$ is one such example.