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This question is from Challenge and Thrill of Pre-College Mathematics.

Prove that $2n^3-3n^2+n$ is divisible by 6 for any natural number $n$.

I know there are many proofs of this using induction and many other ways. Is there any proof using concept of polynomials particularly using remainder theorem and factorization of polynomials? These concepts were introduced just before the exercise and the whole exercise was based on divisibility of a polynomial by another polynomial. Any hint/suggestions are really appreciated. Thank you!

I tried to prove that $2n^3-3n^2+n$ is divisible by $2b_1(n)$ and $3b_2(n)$ where $b_1(n)$ and $b_2(n)$ are polynomials over $\mathbb Z$ but could not found such polynomials.

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  • $\begingroup$ It is easier to show that the value of the polynomial is divisible by both $2$ and $3$ for all $n$ $\endgroup$ – Peter Mar 30 '18 at 7:17
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    $\begingroup$ That polynomial is not divisible by an even polynomial. (Any even multiple of a polynomial has all its coefficients even.) $\endgroup$ – Patrick Stevens Mar 30 '18 at 7:18
  • $\begingroup$ @Peter it is easy using concepts of elementary number theory. Is there any way using polynomials. $\endgroup$ – dssknj Mar 30 '18 at 7:20
  • $\begingroup$ Is $n^2 + n$ always even for $n \in \mathbb{Z}$? Is it divisible by $2b_1(n)$? $\endgroup$ – Eric Towers Mar 30 '18 at 7:25
  • $\begingroup$ I am a little unsatisfied by these answers as these uses number theory but not polynomials as stated OP. $\endgroup$ – dssknj Mar 30 '18 at 8:17
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\begin{align}2n^3-3n^2+n &= n(2n^2-3n+1)\\&=n(2n-1)(n-1)\\ &= (n-1)n(n+1+n-2)\\ &=(n-1)n(n+1)+(n-2)(n-1)n \end{align}

Try to interpret the meaning of the last expression.

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  • $\begingroup$ Oh thank you I was not able to convert it that way. $\endgroup$ – dssknj Mar 30 '18 at 7:24
  • $\begingroup$ Is there any solution without using number theory and just using polynomials? $\endgroup$ – dssknj Mar 30 '18 at 16:15
  • $\begingroup$ Would you like to share anything you know about polynomials that can help you check that a polynomial is divisible by $6$? The moment we talk about divisibility of numbers, I think we are already dealing with number theory. I just uses the properties that between $2$ consecutive integers, there is a multiple of $2$, among $3$ consecutive integers, there is a multiple of $3$. $\endgroup$ – Siong Thye Goh Mar 30 '18 at 16:26
  • $\begingroup$ I am asking this question in this way cause this question is asked in a chapter of polynomials. Now I think it must be a misprint or something. $\endgroup$ – dssknj Mar 30 '18 at 16:43
  • $\begingroup$ This is the best answer here. So I am accepting it. $\endgroup$ – dssknj Apr 2 '18 at 2:10
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The polynomial factors to $$2n^3-3n^2+n=(n-1)n(2n-1)$$

One of the numbers $n-1$ and $n$ must be even and modulo $3$, the factors are $n+2$ , $n$ , $2n+2=2(n+1)$ , so one of them must be divisible by $3$.

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Others have already given a factorization as $(n-1)n(2n-1)$ Now look at what $n$ is modulo 6, then calculate what $(n-1) $ and $2n-1$ are modulo 6 and multiply all three.

n=0 mod 6. Clearly true

n=1 mod 6, then $n-1$ is a multiple of 6 and hence the product is.

n= 2mod 6 then the three numbers are 1,2, and 3 mod 6 and product is 0 mod6.

n=3mod 6, then n-1 is 2 mod6 hence n(n-1) is multiple of 6

n=4mod 6. Then n-1 is 3 mod 6 and so n(n-1) is also a multipe of 6.

n=5mod 6 Then n-1 is 4mod6 and (2n-1) is 3 mod 6, hence the product is 0 mod6

n=3 mod 6 then three numbers are 2,3 and some x, product is 0 mod6

ALTERNATIVE: (a corollary of the formula for sum of squares) Prove the formula $1^2+2^2+\cdots +n^2= n(n+1)(2n+1)/6$

S0 the sum of first $n-1$ squares is $(n-1)n(2n-1)/6$, as this quantity is an integer (sum of integers!) the numerator should be a multiple of 6, the denominator.

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  • $\begingroup$ This answer is using number theory not polynomials. $\endgroup$ – dssknj Mar 30 '18 at 8:15
  • $\begingroup$ Proving something to be a multiple of 6, is a number-theoretic question. So avoiding number theory is an artificial meaningless constraint (often impossible to achieve for this kind of problems. See Eric Tower's comment.) $\endgroup$ – P Vanchinathan Mar 30 '18 at 8:24
  • $\begingroup$ I am not putting that constraint the book is putting that constraint. This question is in chapter 10. Factorization of polynomials and Number theory is chapter 14. $\endgroup$ – dssknj Mar 30 '18 at 9:02

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