Gara Nazionale di Matematica

Solution:

a. We rewrite the equation in the form
$$3^{k}=x^{3}+1=(x+1)\left(x^{2}-x+1\right)$$
Since the only divisors of a power of 3 are themselves powers of 3, every solution must satisfy the system
$$\left\{\begin{array}{l} x+1=3^{a} \\ x^{2}-x+1=3^{b} \end{array}\right.$$
with $a$ and $b$ positive integers such that $a+b=k$. If $a=1$, we get $x=2, b=1$ and $(x, k)=(2,2)$ is a solution. If $a>1$, substituting $x=3^{a}-1$ into the second equation gives
$$3^{2 a}-3^{a+1}+3=3\left(3^{2 a-1}-3^{a}+1\right)=3^{b}$$
which is impossible because $3^{2 a-1}-3^{a}+1$ is greater than 1 and is not divisible by 3. Therefore there is a unique solution, $(x, k)=(2,2)$.

b. Suppose first that $n$ is an odd number. We can write the equation in the form
$$3^{k}=x^{n}+1=(x+1)\left(x^{n-1}-x^{n-2}+\cdots-x+1\right)$$
and, similarly to before, any solution should satisfy the system
$$\left\{\begin{array}{l} x+1=3^{a} \\ x^{n-1}-x^{n-2}+\cdots-x+1=3^{b} \end{array}\right.$$
with $a$ and $b$ positive integers such that $a+b=k$. Dividing the polynomial $x^{n-1}-x^{n-2}+\cdots-x+1$ by $x+1$ we get
$$x^{n-1}-x^{n-2}+\cdots-x+1=q(x)(x+1)+r$$
and, setting $x=-1$, we obtain $r=n$. Hence the greatest common divisor between $x+1$ and $x^{n-1}-x^{n-2}+\cdots-x+1$, which is a power of 3 with positive exponent, must also divide $n$, and hence in particular 3 must be a divisor of $n$. Setting $n=3 m, x^{m}=y$, our equation becomes
$$3^{k}-1=y^{3}$$
From case (a) we know that the only solution is $y=2, k=2$. But $x^{m}=2$ clearly has no integer solutions for $m>1$.

Finally suppose $n=2 m$ is even, and, setting $x^{m}=y$, consider the equation
$$3^{k}=x^{2 m}+1=y^{2}+1$$
Considering all possible remainders of the division of $y$ by 3, namely the cases $y=3 z, 3 z+1, 3 z-1$, we see that $y^{2}+1$ is never divisible by 3, and hence there are no solutions.