# Thread: [SOLVED] induction with multiples

1. ## [SOLVED] induction with multiples

how to prove this

$\displaystyle 5^{2n}-3^n$ when $\displaystyle n\geq0$

i do the base step of proving it works for 0

$\displaystyle 5^{2(0)}-3^0$

$\displaystyle =0$ and $\displaystyle 0\times11=0$ zero is a multiple of 11

so assume $\displaystyle 5^{2n}-3^n$ now the inductive step using n+1. Where do i start?

2. Originally Posted by Rapid_W
how to prove this

$\displaystyle 5^{2n}-3^n$ is a multiple of 11 when $\displaystyle n\geq0$

i do the base step of proving it works for 0

$\displaystyle 5^{2(0)}-3^0$

$\displaystyle =0$ and $\displaystyle 0\times11=0$ zero is a multiple of 11

so assume $\displaystyle 5^{2n}-3^n$ now the inductive step using n+1. Where do i start?
I added the part in bold...that seemed to be a very important part of the problem that was left out.

Now, when you apply the inductive step, suppose for some $\displaystyle k\in\mathbb{N}$ that $\displaystyle 5^{2k}-3^k=11m$ for some $\displaystyle m\in\mathbb{N}$

Now, it follows that

$\displaystyle 5^{2(k+1)}-3^{k+1}=25\cdot5^{2k}-3\cdot 3^{k}=3\left(5^{2k}-3^{k}\right)+22\cdot5^{2k}=3\left(11m\right)+22\cd ot5^{2k}$ $\displaystyle =11\left(3m+2\cdot5^{k}\right)$, which is obviously a multiple of 11.

Does this make sense?

3. yes i can completly follow this thanks