# Find all ordered pairs (m.n) of natural numbers that satisfy the equation

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?

• I think, you've just proved that there is a unique solution. – Tito Eliatron Dec 9 '20 at 19:20
• How is it? can you explain? – lokesh L P Dec 9 '20 at 19:22
• $3^m-1$ 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? – 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.