Exponential diophantine equation: $2p^2-6p+7=3^n$.

Question

I'm trying to prove that the only integer positive solutions are $(n=1,\ p=1)$ and $(n=3,\ p=5)$. Is there a simple way to do that?

Answer 1

Let $p=a$ be any integer. $(2a-3)^2+5=2\cdot 3^n$. Now $3$ cases:

• $n=3k$. Then $(2(2a-3))^2+20=\left(2\cdot 3^k\right)^3$. But $x^2+20=y^3$ has $2$ integral solutions $(x,y)=(\pm 14,6)$ (http://tnt.math.se.tmu.ac.jp/simath/MORDELL/), so $(a,k)=(5,1),(-2,1)$, so $(a,n)=(5,3),(-2,3)$.

• $n=3k+1$. Then $(6(2a-3)))^2+180=\left(2\cdot 3^{k+1}\right)^3$. But $x^2+180=y^3$ has $4$ integral solutions $(x,y)=(\pm 6,6),(\pm 573,69)$, so $(a,n)=(2,1),(1,1)$.

• $n=3k+2$. Then $(18(2a-3))^2+1620=\left(2\cdot 3^{k+2}\right)^3$, no solutions.