# Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$?

Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$?

I tried factoring individual components out, but it seemed really tedious.

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It's easy to get $n\equiv0\pmod2$ and $n\equiv4\pmod5$ but I think the easiest way to find $n$ is to ditch modular arithmetic and just find those 5th powers. –  Gerry Myerson May 3 '12 at 23:50
So why don't you just add them up and take a fifth root? Do you have a purpose? –  Chris K. Caldwell May 3 '12 at 23:52
Since when does math have to have a purpose? –  Mark Adler May 4 '12 at 4:06

If there is such an $n$, it must be a multiple of 6 and 1 less than a multiple of 5, and it must exceed 133 but not by a whole lot, so my money's on 144.
Adding to Gerry's answer, quantifying how much $n$ can exceed $133$ can be done as follows. $n^5 = (27^5 + 84^5 + 110^5 + 133^5) < 133^5 (0.3^5 + 0.7^5 + 0.9^5 + 1)$. Some trivial bounds gives us $0.3^5 < 0.1$, $0.7^5 < 0.2$ and $0.9^5 < 0.7$. Hence, $n^5 < 133^5 (0.1 + 0.2 + 0.7 + 1) = 2 \times 133^5$. Hence, $n$ can be atmost $133 \times 2^{1/5}$ greater than $133$. $2^{1/5} < 2^{1/4} < \sqrt{1.44} = 1.2$. Hence, $134 \leq n \leq 133 \times 1.2$. Also, $n \equiv -1 \mod 5$ and $n \equiv 0 \mod 6 \implies n \equiv 24 \bmod 30$. And, $n = 144$ is the only one in $\{134,\ldots 159\}$. –  user17762 May 4 '12 at 0:34
Further, more strongly, it is very easy to show that $\rm\:n\equiv 1\pmod{13},\:$ viz. $$\rm n^5 \equiv 1^5\!+6^5\!+6^5\!+3^5\equiv 1\!+\!2\!+\!2\!-\!4\equiv 1\:\Rightarrow \rm\:n^5 \equiv 1\equiv n^{12}\:\Rightarrow\: 1 \equiv n^{(5,12)}\! = n$$ –  Bill Dubuque May 4 '12 at 0:37
Thank you! Narrowing down the options makes sense. @Bill, why does $n^5 \equiv 1 \equiv n^{12}$, or more specificially, why is $1 \equiv n^{12}$? –  David Faux May 4 '12 at 15:52
@DavidFaux The calculation shows that, mod $\rm 13,\:$ $\rm\:n^5\equiv 1.\:$ Thus $\rm\:n\not\equiv 0,\:$ so by little Fermat, $\rm\:n^{12}\equiv 1\pmod{13}.\:$ Since $\rm (5,12) = 1\:$ there are integers $\rm\: J,K\:$ such that $\rm\:5J+12K = 1.\:$ Thus $$\rm\ n^1\equiv n^{5J+12K}\equiv (n^5)^J (n^{12})^K\equiv 1^J 1^K \equiv 1\:\ (mod\ 13)$$ More generally, see my posts on order ideals. –  Bill Dubuque May 4 '12 at 17:26