Simply Perfect

Let us consider the equation ${ 3 }^{ 2n }+{ 3n }^{ 2 }+7$ equal to square of a number, suppose ${ x }^{ 2 }$

$\Rightarrow { x }^{ 2 } > {3}^{2n}$

$\Rightarrow {x} > {3}^{n}$

$\Rightarrow {x} \ge {3}^{n}+1$

This gives us 2 equations-

${ 3 }^{ 2n }+{ 3n }^{ 2 }+7 = { x }^{ 2 }$ ...1

AND

${x}^{2} \ge {{3}^{n} + 1} ^ {2}$ ....2

On Equating 1 and 2-

${ 3 }^{ 2n }+{ 3n }^{ 2 }+7\quad =\quad { 3 }^{ 2n }+2{*3 }^{ n }+1$

$\Rightarrow 2.{ 3 }^{ n }\le 3{ n }^{ 2 }+6$

By Hit and Trial, we get that ${n} < {3}$ , Hence ${n} = {1}$ or ${n} = {2}$

If ${n} = {1}$ then, ${ 3 }^{ 2n }+{ 3n }^{ 2 }+7 = 19$ , which is not a perfect square.

Whereas, if ${n} = {2}$ then, ${ 3 }^{ 2n }+{ 3n }^{ 2 }+7 = 100$ which is a perfect square of 10 .

Hence, ${n} = {2}$ is the only solution.