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# Help in Proving Divisibility

While I am proving this statement: An = 5^n + 2 (3^(n-1)) + 1 is divisible by 8 for all positive integers, I was wondering if my method of proof is correct if I use the induction method, then use the theorem that d divides (ax+by) or finding linear combinations, and further proving by parity. Is my method of proof correct for this case? (On need-to-know basis.)

Note by John Ashley Capellan
3 years, 8 months ago

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Yes, induction is the way to go. I would prove it like this:

Basis

Let $$n=1$$. Then we have

$$5^1+2\cdot 3^0+1=1$$,

which is divisible by $$8$$.

Induction step

Let $$5^k+2\cdot 3^{k-1}+1$$ be divisible by $$8$$, for $$k=1,2,\ldots,n$$. Then

$$5^{n+1}+2\cdot 3^n +1$$

$$=5\cdot 5^{n}+3\cdot 2\cdot 3^{n-1} +1$$

$$=4(5^n)+2(2\cdot 3^{n-1})+5^n+2\cdot 3^{n-1}+1$$

$$=4(5^n+3^{n-1})+5^n+2\cdot 3^{n-1}+1$$

From our induction hypothesis, 8 divides $$5^n+2\cdot 3^{n-1}+1$$, and we can therefore write it as $$8c_1$$ for some $$c_1\in \mathbb{Z}$$. Thus, we only need to show, that $$8$$ divides $$4(5^n+3^{n-1})$$, which will be achieved, if we can show that $$2$$ divides $$5^n+3^{n-1}$$ - ie. it is even.

It can be shown that the product of two uneven numbers is uneven, and therefore both $$5^n$$ and $$3^{n-1}$$ are uneven. Furthermore, it can be shown that the sum of two uneven numbers is even. Hence, $$5^n+3^{n-1}$$ is even, and we can write it as $$2c_2$$ for a suitable $$c_2\in \mathbb{Z}$$. We now have

$$=4\cdot 2c_2 + 8c_1=8(c_1+c_2)$$,

which is divisible by 8.

QED · 3 years, 8 months ago