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I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.

bonehead
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    Direct proof: $n^5-5n^3+4n=(n-2)(n-1)(n)(n+1)(n+2)$. $120=2^3\cdot 3\cdot 5$. Among the terms $n-2,n-1,\dots,n+2$ at least one of them will be divisible by five, at least one will be divisible by three, at least one divisible by four and at least one more other than the one divisible by four will also be divisible by two. – JMoravitz Jul 06 '16 at 16:22
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    $n^4+2n^3-n^2-2n = (n-1)n(n+1)(n+2)$. It's divisible by $24$ for $n \in \mathbb{N}$ (you can prove this by induction as well, or simply prove it's divisible by $3$ and by $8$). – Abstraction Jul 06 '16 at 16:24
  • What do you mean k is any whole number? This was for a specific value of n so it should have precisely one value in terms of n. What do you get when you expanded it. – fleablood Jul 06 '16 at 17:15
  • Prove the term in parenthesis us divisible by 24 via induction. You'll get a term multipled by 4 you can prove is divisible by 6. Proving that you get a term multiplied by 3 that you can prove is divisible by 2. Proving that you get a term that that is clearly divisible by 1. You can be perverse and do that. But no-one will blame you if somewhere along the line you decide to simply factor and prove a result directly. Still there's a cute little symmetry in recursive upon recursive induction. Soothing and hypnotic in a way. – fleablood Jul 06 '16 at 20:03
  • All these answers are good but they are all direct. On the other hand induction is very inefficient and repetitive. – fleablood Jul 06 '16 at 20:05

4 Answers4

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Using repeated differences and Newton's interpolation formula we get $$ n^5-5n^3+4n = 120 \binom{n}{3} + 240 \binom{n}{4} + 120 \binom{n}{5} $$ Although this identity suffices for answering the question, it also implies the simpler identity below: $$ n ^5-5n^3+4n = 120 \binom{n+2}{5} $$ which gives a crystal clear answer to the question.

If you must use induction, then: \begin{align} f(n+1)-f(n) &=5 n^4+10 n^3-5 n^2-10 n\\ &= 5 (n+2) (n+1) n (n-1)\\ &= 5 (4!) \binom{n+2}{4}\\ &= 120 \binom{n+2}{4} \end{align}

lhf
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  • Can you explain or link further explanation on how to write polynomials in the form you did? But basically what you did was to prove that every increment to the function was divisible by 120 which in combination with knowing that f(3) is divisible by 120 means that the whole thing is always divisible by 120? – bonehead Jul 07 '16 at 11:36
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    @Ahmed, for the first expression, see the Wikipedia link I gave. The second expression follows directly from Pascal's rule. The last one is exactly what have understood; it depends on $f(3)$ being divisible by $120$. ($f(n)$ for $n=0,1,2$ also works, because $f(n)=0$ in these cases.) – lhf Jul 07 '16 at 11:40
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Let it be true for k

$$k^3(k^2-1) - 4k(k^2-1) = k(k^2-4)(k^2-1) = (k+2)(k+1)k(k-1)(k-2)$$ is divisible by 5,4,3,2 and hence 120

Now replace k by k+1

Then $$(k+3)(k+2)(k+1)k(k-1)$$ is still divisible by 5,4,3,2 and hence 120

It is true for any n>=3.

Satish Ramanathan
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    True. And this is good (better than induction) but technically isn't induction as proving for k+1 in no way uses the property of k. – fleablood Jul 06 '16 at 17:13
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Hint $\ $ Show $\ 3,5,8\mid f(n)\! =\! (n-2)(n-1)n(n+1)(n+2)$

implies that $\ 3,5,8\mid\, f(n+1)\,\ =\,\ (n-1)n(n+1)(n+2)(n+3)$

Bill Dubuque
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  • Further hint: every sequence of $,k,$ consecutive integers contains a multiple of $,k,,$ and $,5\mid n!+!3\iff 5\mid n!-!2\ $ by $\ n!+!3-(n!-!2) = 5\ \ $ – Bill Dubuque Jul 06 '16 at 19:42
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${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$

  • ${n^5-5n^3+4n = (n-2)(n-1)(n)(n+1)(n+2)= 5!\binom{n+2}{5}}$
SSA
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