3

The assertion: $n^3 + 5n$ is divisible by $6$

I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.

This is what I have so far:

For $n=k$: $k^3 + 5k = 6t$

For $n= k+1$:

$(k+1)^3 + 5(k+1)$

$= k^3 + 3k^2 + 8k + 6$

$= (k^3 + 5k) + 3k + 3k^2 + 6$

$= 3k^2 + 3k + 6t + 6$ $= ???$

I cannot pull out a $6$ because that would leave halves that cannot be counted as integers. How should I proceed?

imranfat
  • 10,029
user1861051
  • 357

5 Answers5

6

write $$n^3+5n$$ in the form $$(n-1)n(n+1)+6n$$ and this is the proof.

Dr. Sonnhard Graubner
  • 95,283
3

You can complete your work this way: $$ (k+1)^3 + 5(k+1)=(k^3 + 5k) + 3k + 3k^2 + 6=6t+3k+3k^2+6=6 \left(t+1+\frac{k(k+1)}{2} \right) $$ and note that $k$ or $k+1$ is even.

Emilio Novati
  • 62,675
  • Thank you very much for your help. Continuing with the work I had completed helped me understand what I was missing. – user1861051 Nov 10 '16 at 15:24
2

Hint: Note, $3k+3k^2=3k(k+1)$. As you want to show divisibility by $6$ and $3$ is a factor you only need to show that it is divisible with $2$. What is $k(k+1)$ if $k$ is even? What happens if $k$ is odd?

MrYouMath
  • 15,833
2

Alternatively. Putting $n=k+1$ into your given expression $n^3+5n$ gives $(k+1)^3+5(k+1)$ (Induction) and that expands to $k^3+3k^2+2k+6k+6$ which can be rewritten as $k(k+1)(k+2)+6(k+1)$. Both terms are divisible by $6$ (why?) and from here I essentially concur with Sonnhard

imranfat
  • 10,029
2

Proof by induction

First, show that this is true for $n=1$:

$1^3+5\cdot1=6$

Second, assume that this is true for $n$:

$n^3+5n=6k$

Third, prove that this is true for $n+1$:

$(n+1)^3+5(n+1)=$

$n^3+3n^2+3n+1+5n+5=$

$\color\red{n^3+5n}+3n^2+3n+1+5=$

$\color\red{6k}+3n^2+3n+1+5=$

$6k+3n^2+3n+6=$

$6k+6+3n^2+3n=$

$6(k+1)+3\cdot\color\green{n(n+1)}=$

$6(k+1)+3\cdot\color\green{2m}=$

$6(k+1)+6m=$

$6(k+1+m)$

Please note that the assumption is used only in the part marked red.

Proof by modular arithmetic

Consider the following cases:

  • $n\equiv0\pmod6 \implies n^3+5n\equiv0^3+5\cdot0\equiv6\cdot 0\equiv0\pmod6$
  • $n\equiv1\pmod6 \implies n^3+5n\equiv1^3+5\cdot1\equiv6\cdot 1\equiv0\pmod6$
  • $n\equiv2\pmod6 \implies n^3+5n\equiv2^3+5\cdot2\equiv6\cdot 3\equiv0\pmod6$
  • $n\equiv3\pmod6 \implies n^3+5n\equiv3^3+5\cdot3\equiv6\cdot 7\equiv0\pmod6$
  • $n\equiv4\pmod6 \implies n^3+5n\equiv4^3+5\cdot4\equiv6\cdot14\equiv0\pmod6$
  • $n\equiv5\pmod6 \implies n^3+5n\equiv5^3+5\cdot5\equiv6\cdot25\equiv0\pmod6$

Please note that this method is handy only for a relatively small divisor.

barak manos
  • 43,109