Random problem given to me by nsun

by Wolstenholme, Aug 3, 2014, 11:30 PM

Find all positive integers $n$ and $k$ such that $1^n+2^n+\cdots+n^n=k!$

Solution:

It is clear that $n = 1$ works so assume that $n > 1$ .

Now, assume $n$ is odd. I shall show that $n + 2 \vert 2^n + 3^n + \dots + n^n$ . Grouping terms so that the sum becomes $(2^n + n^n) + (3^n + (n - 1)^n) + ... + \left(\left(\frac{n + 1}{2}\right)^n + \left(\frac{n + 3}{2}\right)^n\right)$ we find since $n$ is odd that $n + 2$ divides each of the terms and so we have the desired result. Therefore since $n + 2 \nmid 1^n + 2^n + \dots + n^n$ we have that $k < n + 2$ but then clearly the RHS is smaller than the LHS so there is no solution if $n$ is a odd integer greater than $1$ .

Now, since we know that $n$ is even, I claim that $v_{2}(1^n + 2^n + \dots + n^n) = v_{2}(n) - 1$ . To prove this, consider the identity $1^n + 2^n + \dots + n^n = (1^n - 1) + (2^n) + (3^n - 1) + (4^n) + \dots ((n - 1)^n - 1) + (n^n) + \frac{n}{2}$ . By LTE, each of the terms in parenthesis is divisible by $2^{v_{2}(n)}$ which is obviously greater than $v_{2}(n) - 1$ . But $v_{2}\left(\frac{n}{2}\right) = v_{2}(n) - 1$ so my claim is proven. But since $k > n$ we have that $v_{2}(k!) > v_{2}(n) - 1$ which means that equality cannot hold.

Therefore the only answer is $n = 1$ .

This post has been edited 1 time. Last edited by Wolstenholme, Aug 4, 2014, 12:22 AM

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
this blog is so hot
by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
by User360702, Jan 10, 2019, 6:03 PM
helloooooo
by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
You were just featured on AoPS's facebook page.
by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

18 shouts

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Random problem given to me by nsun

by Wolstenholme, Aug 3, 2014, 11:30 PM

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