What is the positive integer value of $n$ if $3^{n} + 81$ is a perfect square?

[juantheron]

If $3^{n} +81$ is a perfect square, find a positive integer value of $n$.

My Trail:: When $n\leq 4,$ then easy to know that $3^{n} +81$ is not a perfect square.

Now let $\displaystyle n = k +4 (k\in \mathbb{Z^{+}}),$ then $3^{N} +81 = 81 (3^{k} +1).$

So $3^{N} +81$ is a perfect square, and $81$ is square,

there must be a positive integer $x,$ such that

$3^{k}+1 = x^2\Rightarrow 3^k = (x-1)\cdot (x+1)$

Now How can i solve after that

Help me

Thanks

 

[MarkFL]

I would write:

\(\displaystyle 3^n+81=m^2\)

\(\displaystyle 3^n=(m+9)(m-9)\)

Now, observing that:

\(\displaystyle 18+9=27=3^3\) and \(\displaystyle 18-9=9=3^2\)

What value do we obtain for $n$?

 

[mente oscura]

If $3^{n} +81$ is a perfect square, find a positive integer value of $n$.

Hello.

[tex]3^n+81[/tex], It cannot be a perfect square, For being, both, odd numbers.

Demostration:

[tex](2n-1)^2+(2m-1)^2=4(n^2+m^2)-4(n+m)+2=[/tex]

[tex]=2[2(n^2+m^2)-2(n+m)+1][/tex]

The square root, of the latter expression, has to be an irrational number, for being divisible, only once for "2".

Regards.

 

[mathbalarka]

$3^n+81$, It cannot be a perfect square

False. I can find a precise value $n$ for which the above is a perfect square.

 

[mathbalarka]

Spoiler

Hint : Prove that $18$ is divisible by $m - 9$.

 

[mente oscura]

False. I can find a precise value $n$ for which the above is a perfect square.

Hello.

I am sorry. Really, I have considered "n", even number and, it is not possible, but yes it can be an odd number.

Regards.