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For how many $n$ is $2021^n$ + $2022^n$ + $2023^n$ + ... + $2029^n$ prime?

My first thought is set x = 2021 so we can create:

$x^n$ + $(x+1)^n$ + $(x+2)^n$ + ... + $(x+8)^n$

And then we expand each term even though we don't know $n$?

So for example: $(x+1)^n$ = $x^n$ + $\dbinom{n}{1}$$x^{n-1}$ + $\dbinom{n}{2}$ $x^{n-2}$ + ... + $\dbinom{n}{n-1}$ $x^{n-(n-1)}$ + $\dbinom{n}{n}$ $x^{n-n}$

I understand this, but how do I go about showing that the original sum is prime?

I don't think some obvious pattern will arise when I've expanded every term that will show whether or not it is prime.

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  • $\begingroup$ I don;t know how to prove it, but Wolfram Alpah seems to say that the value of that expression is divisible by $5$ when $n\neq 4k$ and divisible by $3$ when $n=4k$ $\endgroup$
    – Sayan Dutta
    Commented Sep 7, 2021 at 21:46
  • $\begingroup$ A good idea when working with consecutive numbers is to take the middle term here $x=2025$ then you have $(x-4)^n+(x-3)^n+\cdots+(x+3)^n +(x+4)^n$, now in this case it's not that useful though it makes it sort of obvious that for odd $n$ the expression is divisible by $x$. $\endgroup$
    – kingW3
    Commented Sep 7, 2021 at 23:40

1 Answer 1

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Note that $x^n$, $(x+3)^n$ and $(x+6)^n$ give the same remaider when divided by $3$. So, the sum $x^n+(x+3)^n+(x+6)^n$ must be divisible by $3$.

Now, the expression in the question can be written as $$(2021^n+2024^n+2027^n)+(2022^n+2025^n+2028^n)+(2023^n+2026^n+2029^n)$$ which proves that this sum is divisible by $3$.

So, none of those expressions are prime.

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