Use modular arithmetic to prove that for $n \geq 1$, the quantity $2^{3n+1}+ 3 \times 5^{2n+1}$ is always divisible by 17.
Prove 17 divides $2^{3n+1}+ 3 \times 5^{2n+1}$
3
$\begingroup$
elementary-number-theory
discrete-mathematics
-
1You should add that $n\in\mathbb Z^+$, or $n$ is an integer or a natural number. – 2017-01-01
3 Answers
4
$$2^{3n+1}+3\cdot5^{2n+1}=2\cdot(2^3)^n+3\cdot5\cdot(5^2)^n$$
Now observe that $5^2\equiv2^3\pmod{17}$
1
$$2^{3n+1}+3\cdot5^{2n+1}=17\cdot8^n+15(25^n-8^n)$$ and we are done!
0
It's easy to verify as in lab's answer. Below we show how to discover it.
It boils down to $\ 3(\color{#c00}3^{\large\color{#c00}-3})^{\large 2N+1}\!\! =\! (\color{#c00}3^{\large\color{#c00} -2})^{\large 3N+1}\, $ in modular form, $ $ e.g.
$\ \ \ {\rm mod}\ \ \color{#0a0}{3a\!-\!1}\!: \ \ 3(\color{#c00}a^{\large 3})^{\large 2N+1} \equiv\ (\color{#c00}a^{\large 2})^{\large 3N+1} \ {\rm by}\ \ \color{#c00}{a\equiv 3^{-1}} \ {\rm by}\ \ \color{#0a0}{3a\equiv 1} $
$ \begin{align} {\rm e.g.}\ \ &{\rm mod}\ \ \ \ 5\!:\,\ 3\,\cdot\, 3^{\large 2N+1} \equiv\ \ 4^{\large 3N+1}\\ &{\rm mod}\ \ \ \ 8\!:\,\ 3\,\cdot\, 3^{\large 2N+1} \equiv\ \ 1\\ &{\rm mod}\,\ 11\!:\,\ 3\,\cdot\, 9^{\large 2N+1}\equiv\ \ 5^{\large 3N+1}\\ &{\rm mod}\,\ 14\!:\,\ 3\cdot 13^{\large 2N+1}\!\equiv 11^{\large 3N+1}\\ &{\rm mod}\,\ 17\!:\,\ 3(-5)^{\large 2N+1}\!\!\equiv\ 2^{\large 3N+1}\quad \text{[the OP]}\\ \end{align}$