Find all ordered pairs $(m, n)$ of natural numbers that satisfy the equation

$9^m +3^m-2 = 2 p^n$

Where $p$ is a prime number.

This is what I have tried. I wrote the equation as $ (3^m+2)(3^m-1) = 2p^n$

$(3^m-1)$ is always even and $(3^m+2)$ is always odd. So $p$ cannot be even i.e. $p$ cannot be $2$.

When $(3^m-1) = 2 \implies m = 1.$ Then $p=5$ and $n=1$. $(1,1)$ is one ordered pair.

I feel there are no other solutions but I don't know how to prove it. Can someone help?

  • $\begingroup$ I think, you've just proved that there is a unique solution. $\endgroup$ – Tito Eliatron Dec 9 '20 at 19:20
  • $\begingroup$ How is it? can you explain? $\endgroup$ – lokesh L P Dec 9 '20 at 19:22
  • 1
    $\begingroup$ $3^m-1$ need not be only 2 but it can be any number which is twice of a prime number, p. Say when m=3, $3^m-1$ = 26 = 2*13. In this case $3^m+2$ is not 13 or any power of 13. so, m=3 is not a solution. How to prove that for not other value of m, the equation is true? $\endgroup$ – lokesh L P Dec 9 '20 at 19:31

If there is any other solution, where m is not equal to 1 then observe that both factors necessarily have gcd as $p^k$ where k is a positive integer. But this is impossible since their difference is 3. Since you showed uniquness for m=1, you are done.


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