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If $3^n+81$ is a perfect square, Then calculation of a positive integer value of $n$.

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

Now let $n=k+4(k∈Z^{+}),$ 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 $3^{k}+1=x^2⇒3^k=(x−1)⋅(x+1)$

Means $(x+1)$ and $(x-1)$ must be a power of $3$ form

Now I did not understand how can i solve after that

Help Required

Thanks.

juantheron
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3 Answers3

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Note that if $3^n+81 = x^2$, we have $$(x+9)(x-9) = 3^n$$ Hence, we have $x+9 = 3^m$ and $x-9 = 3^{n-m}$. Hence, we need two powers of $3$ that differ by $18$, i.e., we need $3^m - 3^{n-m} = 18$.

Now observe that $3^m-3^{m-1} \geq 18$, if $m\geq 3$. Hence, we have limited options for $m$ and $n-m$.

Hence, $m=3$ and $n-m=2$ and we are done.

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You are finished. Your condition forces $x=2$. Other pairs of powers of $3$ are separated by much more than $2$.

André Nicolas
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  • Thanks André Nicolas, but I did not understand the line Other pairs of powers of 3 are separated by much more than 2.please explain me – juantheron Dec 05 '13 at 03:33
  • The powers of $3$ are $1,3,9,27,\dots$. Our two numbers $x-1$ and $x+1$ are both powers of $3$, and their difference is $2$. If we look at the table of powers of $3$, only $3$ and $1$ differ by $2$, other differences are far large (the next one is $6$.) – André Nicolas Dec 05 '13 at 04:05
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$(x-1)$ and $(x+1)$ must be power of 3. Then $x-1=3^a$ and $x+1=3^b$

$\Rightarrow 3^b-3^a = 2 \Rightarrow 3^a(3^{b-a}-1)=2$.

So $a=0$ and $b-a=1 \Rightarrow b=1$. Thus $x-1 = 3^a = 1$ and $x+1 = 3^b = 3$.

$\Rightarrow 3^k=1\cdot 3=3 \Rightarrow k = 1 \Rightarrow n=k+1=5$.

Therefore, the unique solution is $n=5$ with $3^5+81=324 = 18^2$.

Tiago Emilio Siller
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