Proving divisibility for $256 \mid 7^{2n} + 208n - 1$

Question

I can't come up with a way of proving this:

$$256 \mid 7^{2n} + 208n - 1\\ \forall n \in \Bbb N$$

I've tried by induction but couldn't see when to apply the inductive hypothesis...

$$P(n+1) = 7^{2n+2}+208n+207$$

How can I continue? Thank you

Answer 1

By inductive hypothesis $7^{2n}+208n-1=256m$, $m\in \mathbb{N}$

Therefore $7^{2n+2}+208n+207=49(256m+1-208n)+208n+207=49\cdot256m-48\cdot208n+256=49\cdot256m-39\cdot256n+256=256(49m-39n+1)$ is divisible by $256$.

Answer 2

By the binomial theorem, $7^{2n}=49^n=(48+1)^n=48^2a+48n+1$ and so $$7^{2n} + 208n - 1 = 48^2a+256n = 256(9a+n)$$

Answer 3

We have transforming step by step suitably, $$7^{2n}+208n-1=$$ $$(48+1)^n+(256-48)n-1=$$ $$(2^4\cdot3+1)^n+(256-2^4\cdot3)n-1 =$$ $$\sum_{k=0}^{k=n-1}\binom nk (2^4\cdot3)^{n-k}+1+256n-2^4\cdot3n-1=$$ $$=\sum_{k=0}^{k=n-2}\binom nk (2^4\cdot3)^{n-k}+256n+2^4\cdot3n-2^4\cdot3n=$$ $$=\sum_{k=0}^{k=n-2}\binom nk (2^4\cdot3)^{n-k}+256n\equiv 0\pmod{256}$$

Answer 4

Hint $\,\ {\rm mod}\ 16^2\!:\ 7^{2n} \equiv\, 48n+1\,\ $ [i.e. $\,P(n)$], $\,$ when multiplied by $\,7^2$

implies $\,$ that $\ \ 7^{2(n+1)}\equiv \color{#0a0}{7^2}(\color{#0a0}{48}n+1)\equiv \color{#c00}{48}n\!+\!49 = 48(n\!+\!1)+1\,\ $ [i.e. $P(n\!+\!1)$]

because $\,\ \color{#0a0}{7^2(48)} = (1+3\!\cdot\! 16)(3\!\cdot\!16)\equiv \color{#c00}{48}\ $ by $\ 16^2\equiv 0\,\ $ using only mental arithmetic!