Question

Sn = n² + 20n +12 20 2 (n +10) ² -88
n belongs to positive integer
Sn is a perfect square
Find the how values of n will satisfy this?

Answer 1

Given:- $S(n)=n^{2}+20 n+12\\=n^{2}+20 n+100-88\\=(n+10)^{2}-88\\\text { Let } n+10=a$

$\text { We now need to solve } a^{2}-88=m^{2} \text { in the integers. }\\\begin{array}{l}\Longrightarrow a^{2}-m^{2}=88 \\(a+m)(a-m)=88=2^{3} \times 11\end{array}\\\text { Let } a+m=k ; \text { where } k \text { is any factor of } 88\\\begin{array}{l}\Longrightarrow=a-m=\frac{88}{k} \\\Longrightarrow a=\frac{k}{2}+\frac{44}{k} ; m=\frac{k}{2}-\frac{44}{k}\end{array}$

$\text { Because } a \text { and } m \text { must both be integers, } k \text { must be even and } k \text { cannot be } 88 \text {. }$$\text { Thus, the possible values of } k \text { are } 2,4,8,22 \text { and } 44 \text {. }\\\text { However, note that } k=a+m=n+10+m>10\\\Longrightarrow k=22 \text { or } 44\\\text { If } k=22 \text {, }\\\begin{array}{l}a=\frac{22}{2}+\frac{44}{22}=11+2=13 \\\Longrightarrow n=a-10=13-10=3 \\\text { If } k=44, a=\frac{44}{2}+\frac{44}{44}=22+1=23 \\\Longrightarrow n=a-10=23-10=13\end{array}$