# Can $7^n+1$ be a perfect cube when $n\geq2$?

$$7^n+1$$ (when $$n\geq2$$) can in principle be a perfect cube, since it sometimes is divisible by $$8$$, and sometimes leaves $$8$$ as a remainder when dividied by $$9$$. But when I tried to find perfect cubes that is form of $$7^n+1$$, I did not suceed.

Can $$7^n+1$$ be a perfect cube?

• Please use MathJax. Here is a tutorial. For $2^n+1$ we have a nice solution...and in general it follows from "Catalan". Jan 27, 2023 at 10:30
• It is known that the only pair of consecutive perfect powers is $(8/9)$. But maybe we can show that this cannot be a cube easier. Jan 27, 2023 at 10:32

This solution is based on a solution to a similar problem about $$2^n + 1$$ found here.

No.

Suppose that $$7^n + 1 = m^3$$ for some positive integers $$n$$ and $$m$$. Rearranging, we get $$7^n = m^3 - 1 = (m-1)(m^2 + m + 1).$$

Because $$7$$ is prime, we know that both $$m-1$$ and $$m^2 + m + 1$$ must both be powers of $$7$$ (including $$7^0 = 1$$).

• If $$m-1 = 7^0 = 1$$, then $$m = 2$$ and hence $$n = 1$$.
• If $$m-1 \neq 7^0$$, then $$7 \mid (m-1)$$. This implies $$m \equiv 1 \pmod 7$$. However, we would get $$m^2 + m + 1 \equiv 1^2 + 1 + 1 \equiv 3 \pmod 7.$$ This contradicts the fact that $$m^2 + m + 1$$ must be a power of $$7$$.

So, excluding $$n = 1$$, there is no other positive integers $$n$$ such that $$7^n + 1$$ is a perfect cube.

• Congratulations! I like your answer very much. Jan 27, 2023 at 12:43
• @VTand You might be interested in looking at Problem 19 for Baltic Way 2022: balticway2022.no/problems. Aug 6, 2023 at 12:52