Find the smallest positive integer n ≥ 10 such that n+6 is a prime...

Since $9 n+7$ is a perfect square Let us assume that $9 n+7=\mathrm{m}^2$ Also $n+6$ is a prime number $\Rightarrow n+6$ must be odd number $\Rightarrow \mathrm{n}$ also must be an odd number So Let us assume $\mathrm{n}=2 \mathrm{k}+1$....(ii) where k is an Integer $$\begin{aligned} & 9(2 k+1)+7=m^2 \\ & \Rightarrow 18 k=m^2-16 \\ & \Rightarrow 18 k=(m+4)(m-4) ...(iii) \end{aligned}$$ Since $18 k$ is even and $m+4$ and $m-4$ both are of some parity $\Rightarrow m$ must also be even. say $m=2 p$, where $p$ is an Integer Substituting in (iii) $$\begin{aligned} & 18 k=(2 p+4)(2 p-4)=4(p+2)(p-2) \\ & \Rightarrow 9 k=2(p+2)(p-2) \end{aligned}$$ $\Rightarrow \mathrm{k}$ must be even, say $\mathrm{k}=2 \ell$ where $\ell$ is an Integer $$\begin{aligned} & 18 \ell=2(\mathrm{p}+2)(\mathrm{p}-2) \\ & \Rightarrow 9 \ell=(\mathrm{p}+2)(\mathrm{p}-2) \end{aligned}$$ $\Rightarrow p$ must be of the form $9 q+2$ or $9 q-2$, where $q$ is an Integer First take $p=9 q-2 \Rightarrow m=2(9 q-2)$ Take $\mathrm{q}=1, \mathrm{~m}=2 \times 7=14 \Rightarrow \mathrm{~m}^2=196$ $9 n+7=196 \Rightarrow 9 n=189 \Rightarrow n=21$ $\Rightarrow n+6=27$ which is not prime Next take $p=9 q+2 \Rightarrow m=2(9 q+2)$ Take $\mathrm{q}=1 \Rightarrow \mathrm{~m}=2 \times 11=22 \Rightarrow \mathrm{~m}^2=484$ $9 n+7=484 \Rightarrow 9 n=477 \Rightarrow n=53$ $\Rightarrow n+6=59$ which is a prime So $\mathrm{n}=53$ is the smallest such number