# Thread: Divisible by 7

1. I might as well add to this.
Suppose that it is true for $\displaystyle K$, i.e. $\displaystyle 3^{2K+1}+2^{K+2}}$ is divisible by $\displaystyle 7$.
Look at $\displaystyle 3^{2(K+1)+1}+2^{(K+1)+2}=3^{2K+3}+2^{K+3}$
So $\displaystyle 3^{2K+3}+2^{K+3}=3^{2K+3}+3^2 \cdot 2^{K+2}-3^2\cdot 2^{K+2}+2^{K+2}$$\displaystyle =3^{2}\left(3^{2K+1}+2^{K+3}\right)-2^{K+2}\left(3^2-2\right)$.
That is clearly divisible by $\displaystyle 7$.

2. Originally Posted by Plato
I might as well add to this.
Suppose that it is true for $\displaystyle K$, i.e. $\displaystyle 3^{2K+1}+2^{K+1}}$ is divisible by $\displaystyle 7$.
I think you've meant $\displaystyle 3^{2K+1}+2^{K+2}}$...

3. Originally Posted by Soroban
Hello, yuud!

Here is a sneaky method . . . if you know modulo arithmetic.

$\displaystyle 3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}$

. . . . . . . . . . .$\displaystyle =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; -1 + 1\text{ (mod 7)}$

$\displaystyle 3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for all positive intergers $\displaystyle n$.

I haven't studied modular arithmetic... but THANKS

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