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Everything is ok except for when I try to substitute $n = k + 1$. I get $8J + 4k + 8$ (where J is an integer) and cannot further factorize out an 8 to prove it is divisible. Is this even a correct substitution?

user71207
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3 Answers3

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The case for $n=2$ is clear. Then we we must show that if the expression is true for some $n$ then it is true for $n+2$. For some $n$ we know that $8|n(n+2) $and we must show that from this information $8|(n+2)(n+4)$. Notice that $(n+2)(n+4) = n(n+2) + 4(n+2)$. Since $n $is even, $n = 2k$ so that $4(n+2) = 4n + 8= 8k + 8$ which $8$ divides. Then $8|n(n+2)$ and $8|4(n+2)$ so that $8|(n+2)(n+4).$

Derek Luna
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If $n$ is even, then $n = 2m$ for some $m$.

Then $n(n+2) =2m(2m+2) =4m(m+1)$.

Now do your induction on $m$, not $n$.

marty cohen
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    Pointless as obviously $2$ divides $m(m+1). $Probably requires induction on the same expression from the start. – Derek Luna Nov 11 '20 at 02:31
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If $n$ is even, then both $n$ and $n + 2$ are even (i.e, divisible by 2). If $n$ is also divisible by $4$, then $n(n + 2)$ is divisible by $4\cdot 2 = 8$. However, if $n$ is not divisible by $4$, then $n$ must have a remainder of $2$ when divided by $4$, which means $n+2$ is divisible by $4$. Then, $n(n+2)$ is divisible by $2\cdot 4 = 8$.

Thus, $\boxed{n(n+2)\text{ is always divisible by }8.}$

Joshua Wang
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