2

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.

David Faux
  • 3,425

2 Answers2

10

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.

Gerry Myerson
  • 179,216
  • 3
    Your money's safe. :-) – Brian M. Scott May 04 '12 at 00:09
  • 1
    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 ${134,\ldots 159}$. –  May 04 '12 at 00:34
  • 4
    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 04 '12 at 00:37
  • 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 04 '12 at 15:52
  • 2
    @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 04 '12 at 17:26
0

Tabulating the expression with respect to low primes:

$\bmod 2: 27^5 + 84^5 + 110^5 + 133^5 \equiv 1^5 + 0^5 + 0^5 + 1^5 \equiv 0 \implies n\equiv 0$

$\bmod 3: 27^5 + 84^5 + 110^5 + 133^5 \equiv 0^5 + 0^5 + -1^5 + 1^5 \equiv 0 \implies n\equiv 0$

$\bmod 5: 27^5 + 84^5 + 110^5 + 133^5 \equiv 2^5 + (-1)^5 + 0^5 + (-2)^5 \equiv -1 \implies n^5 \equiv n\equiv -1$

Collecting these gives $n\equiv 24\bmod 30$, which already points at $144$

$\bmod 7: 27^5 + 84^5 + 110^5 + 133^5 \equiv (-1)^5 + 0^5 + (-2)^5 + 0^5 \equiv -1+3 \equiv 2 \equiv n^5\equiv n^{-1} \implies n \equiv 4$

This confirms that $144$ is the only possible solution in range; the next modular equivalence solution would be far out of range at $144+210 = 354$.

Joffan
  • 39,627