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
7 years, 9 months ago

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Comments

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

Basis

Let n=1n=1. Then we have

51+2⋅30+1=15^1+2\cdot 3^0+1=1,

which is divisible by 88.

Induction step

Let 5k+2⋅3k−1+15^k+2\cdot 3^{k-1}+1 be divisible by 88, for k=1,2,…,nk=1,2,\ldots,n. Then

5n+1+2⋅3n+15^{n+1}+2\cdot 3^n +1

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

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

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

From our induction hypothesis, 8 divides 5n+2⋅3n−1+15^n+2\cdot 3^{n-1}+1, and we can therefore write it as 8c18c_1 for some c1∈Zc_1\in \mathbb{Z}. Thus, we only need to show, that 88 divides 4(5n+3n−1)4(5^n+3^{n-1}), which will be achieved, if we can show that 22 divides 5n+3n−15^n+3^{n-1} - ie. it is even.

It can be shown that the product of two uneven numbers is uneven, and therefore both 5n5^n and 3n−13^{n-1} are uneven. Furthermore, it can be shown that the sum of two uneven numbers is even. Hence, 5n+3n−15^n+3^{n-1} is even, and we can write it as 2c22c_2 for a suitable c2∈Zc_2\in \mathbb{Z}. We now have

=4⋅2c2+8c1=8(c1+c2)=4\cdot 2c_2 + 8c_1=8(c_1+c_2),

which is divisible by 8.

QED

René Christensen - 7 years, 9 months ago
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