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.)

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

BasisLet \(n=1\). Then we have

\(5^1+2\cdot 3^0+1=1\),

which is divisible by \(8\).

Induction stepLet \(5^k+2\cdot 3^{k-1}+1\) 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 – René Christensen · 3 years, 8 months ago

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