prove that the $9th$ power of any integer is of the form $19k$ or $19k \pm 1$
For small power $p$ when the integer is of the form $nk$ or $nk \pm 1$. we use division algorithm $a=nk+r$ then break down $a^n$ and put value of remainder $0$ to $n-1$ and find form of the integer $a$.But for big value of $n$ the process is too labours .Can anyone give me any simpler method??
when $a=19c$, $\space a^9=19^9c^9$ so $a^9\equiv0 (mod19)$ $\Rightarrow a^9=19k$
when $x\neq 19c$ using Fermat's little theorem $a^{18}\equiv 1(mod19)$ so $a^9\equiv \pm1(mod19)$ $\Rightarrow a^{9}=19k\pm1$
Therefore $a^{9}=19k$ or $19k\pm1$
For any integer $n=19k+a,\ 0\le a<19$ $$n^9 = (19k + a)^9 = (19k)^9 + 19a(19k)^8 + 171a²(19k)^7 + .... + 171(a^7)(19k)² + 19(a^8)(19k) + a^9 $$
Now see that we can factor $19$ out of the $1^{st}$ $19$ terms, so we only have to show $a^9, 0\le a < 19$ is of the required form.
$$0^9 = 0 = 19(0)$$ $$1^9 = 1 = 19(0) + 1$$ $$2^9 = 512 = 19(27) - 1$$ $$3^9 = 19683 = 19(1036) - 1$$ we can get $4^9$ by squaring $2^9:$ $$[ 19(27) - 1]^2 = 19^2(27)^2 - 2(19)(27) + 1$$ So, $$4^9 = 19[ 19(27)^2 - 2(27)] + 1 $$ $$6^9 = (2^9)(3^9) = 19[ 19(27)(1036) - 27 - 1036)] + 1 $$ similarly we can get $8^9, 9^9, 12^9, 16^9, 18^9$
once we have $12^9,$ we get $7^9,$ from $(19 - 12)^9 = 9$ multiples of $19 - 12^9$
then we can get $14^9,$ then $5^9$ from $(19 - 14)^9,$ and so on.
$17^9 = (19 - 2)^9 = 8$ multiples of $19,$ say $19k, - 2^9,$ so that's $19k - ( 19(27) - 1) = 19(k-27) + 1$
