$n \in \mathbb{N} \ 5|\ 2^{2n+1}+3^{2n+1}$

Question

show for all $n \in \mathbb{N}$, $$5|\ 2^{2n+1}+3^{2n+1}$$

Indeed,

we've to show that : $2^{2n+1}+3^{2n+1}=0[5] $

note that $2^{2n+1}+3^{2n+1}=2.4^n+3.9^n= $

Answer 1

You're basically there!! Now work modulo 5:

$2^{2n+1} + 3^{2n+1} = 2.4^n + 3.9^n \equiv 2.(-1)^n + 3.(-1)^n \equiv 5.(-1)^n \equiv 0$ (mod $5$)

Answer 2

You can do it by induction. For $n=0$, we have $$2^{2n+1}+3^{2n+1}=2+3=5$$ which is divisible by $5$. Now suppose that $2^{2k+1}+3^{2k+1}$ is divisible by $5$, i.e. $$\tag{1}2^{2k+1}+3^{2k+1}=5N$$ for some integer $N$. Now consider $$2^{2(k+1)+1}+3^{2(k+1)+1}=4\cdot 2^{k+1}+9\cdot3^{2k+1}= 4\cdot (2^{k+1}+3^{2k+1})+5\cdot 3^{2k+1}=5(N+3^{2k+1})$$ where the last equality follows from $(1)$, which is divisible by $5$.

Answer 3

Its much simpler, $3\equiv -2 \pmod 5$. So $$2^{2n+1} +3^{2n+1}\equiv 2^{2n+1} +(-2)^{2n+1} =2^{2n+1}-2^{2n+1}=0 \pmod 5$$

Answer 4

Hint $\,k\,$ odd $\,\Rightarrow a\!+\!b \mid a^k\!+b^k$ by Factor Theorem, or by $\,{\rm mod}\ a\!+\!b\!:\ a\equiv -b\,\Rightarrow\, a^k \equiv (-b)^k \equiv -b^k$

Answer 5

Write $2^{2n+1}+3^{2n+1} = 2^{2n+1}+(5-2)^{2n+1}$ and note that $(5-2)^{2n+1}$ expands to $5t-2^{2n+1}$.