Find all positive integers $n$ so that there are at least two coefficients in the expansion of $(7x+1)^n$ are equal.
I worked out the special case that two "consecutive coefficients" are equal and get $n=8k-1$. Not sure how to do with general cases. Also, I would like to know can we generalize the question to $(px+q)^n$ or something? Thanks in advance.
Edit: What I have done: Suppose there are two consecutive terms with same coefficients. Say $7^m\binom{n}{m}=7^{m+1}\binom{n}{m+1}$. Simplifying the relation yields $m+1=7(n-m)$ or $7n=8m+1$. This holds if and only if $n$ takes the form $8k-1$.
For general cases, I tried to modify the equation to $7^m\binom{n}{m}=7^{m+r}\binom{n}{m+r}$ and it turns out that I need the product of $r$ consecutive integers equals to another product of $r$ consecutive integers multiplied by $7^r$. Then I stuck here.
Edit: I used python to check that if $n=8k-1$ is true. No counterexamples in $n<3000$. It suggests that only $n=8k-1$ works.