Application of Method of Differences

How would you apply the Method of Differences to solve this problem:

f(x)f(x) is a polynomial of degree 2 such that that

f(1)f(2),f(2)f(3),f(3)f(4). f(1) | f(2), f(2) | f(3), f(3) | f(4).

Show that f(2)=±f(1) f(2) = \pm f(1) .

Note by Calvin Lin
6 years, 3 months ago

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Let g(n)=f(n+1)f(n),h(n)=g(n+1)g(n)g(n)=f(n+1)-f(n), h(n)=g(n+1)-g(n). Then g is a polynomial of degree 1, and h is a polynomial of degree 0, i.e. a constant.

Clearly from the given condition, f(2)g(2)f(2)|g(2) and f(3)g(3)f(3)|g(3). Hence f(2)g(3)f(2)|g(3), so f(2)h(2)f(2)|h(2). But h(1)=h(2)h(1)=h(2), so f(2)h(1)=g(2)g(1)f(2)|h(1)=g(2)-g(1), so f(2)g(1)=f(2)f(1)f(2)|g(1)=f(2)-f(1), so f(2)f(1)f(2)|f(1).

Since f(1)f(1) and f(2)f(2) divide each other, they must have the same absolute value, which was what we wanted.

Ang Yan Sheng - 6 years, 3 months ago

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That's a great solution. I tried expressing f(2)f(2) as some factor times f(1)f(1) and f(3)f(3) as some factor times f(2)f(2), but I just got a big mess.

Tim Vermeulen - 6 years, 3 months ago

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You can see how knowing the "method of differences" makes this proof seem almost immediate / obvious.

I would encourage you to work through your "big mess", especially in light of Ang's solution above. Use your notation to work through, see how it follows through, and why you weren't able to push through.

Calvin Lin Staff - 6 years, 3 months ago

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@Calvin Lin Thanks for the suggestion. In fact it did work out quite nicely, although my solution is still not as slick as Ang's. Here it is:

Let f(2)=af(1),f(3)=bf(2),f(4)=cf(3).f(2) = a \cdot f(1), f(3) = b \cdot f(2), f(4) = c \cdot f(3). Then using simple algebra, I obtain

{D2(1)=f(1)(ab2a+1),D2(2)=f(1)(abc2ab+a). \begin{cases} D_2(1) = f(1) \cdot (ab -2a +1), \\ D_2(2) = f(1) \cdot (abc -2ab+a). \end{cases}

As D2(n)D_2(n) is constant for all nn, those two expressions are the same:

f(1)(ab2a+1)=f(1)(abc2ab+a). f(1) \cdot (ab -2a +1) = f(1) \cdot (abc -2ab+a).

Division by af(1) a \cdot f(1) is allowed, as the conditions imply a,f(1)0 a,f(1) \not = 0:

b2+1a=bc2b+1. b-2 + \frac{1}{a} = bc - 2b +1.

The right hand side is an integer, so the left hand side is too: this implies 1a \frac{1}{a} is an integer as well, so a{1,1}a\in \{ -1,1 \}.

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen This looks good.

However, why do you say that "the conditions imply a,f(1)0a, f(1) \neq 0"? How do the conditions imply that? Why can't either of them be 0?

Calvin Lin Staff - 6 years, 3 months ago

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@Calvin Lin Well, f(1)f(2)f(1) \mid f(2), and 00 doesn't divide anything, so f(1)0f(1) \not = 0. Similarly, f(2)f(3)    f(2)0f(2) \mid f(3) \implies f(2) \not = 0. And as a=f(2)f(1)a=\frac{f(2)}{f(1)}, aa is not 00 either.

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen Rethink that statement. Review the definition of | , which was given below.

In particular, Does 0k 0 \mid k or k0 k \mid 0 ?

Calvin Lin Staff - 6 years, 3 months ago

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@Calvin Lin That's interesting. I did think about whether 000 \mid 0, but I figured that didn't work, as 00\frac{0}{0} is undefined. But on the other hand, there is an integer aa (there are plenty!) such that a0=0a \cdot 0 = 0. This doesn't completely destroy my argument, though, but it needs some refinement:

For the sake of contradiction, let's assume that f(1)=0f(1) = 0. Then f(1)f(2)    f(2)=0f(1) \mid f(2) \implies f(2) = 0, because for every integer aa, a0=0a \cdot 0 = 0. Similarly, f(2)f(3)    f(3)=0f(2) \mid f(3) \implies f(3) = 0. So, f(1)=f(2)=f(3)=0f(1)=f(2)=f(3)=0. But f(x)f(x) is a second degree polynomial, so that is impossible. Therefore, the assumption that f(1)=0f(1)=0 was wrong, so f(1)0f(1) \not = 0. Using a similar argument (if I assume that f(2)=0f(2)=0), f(2)f(3)    f(3)=0f(2) \mid f(3) \implies f(3) = 0 and f(3)f(4)    f(4)=0f(3) \mid f(4) \implies f(4) = 0, which is not possible due to the degree of f(x)f(x), so f(2)0f(2) \not = 0. Finally, assuming a=0a=0, f(2)=af(1)    f(2)=0f(2) = a \cdot f(1) \implies f(2)=0, but f(2)0f(2) \not = 0, so a0a \not = 0.

I have made an assumption here: if f(x)f(x) is a second degree polynomial, then f(1)=f(2)=f(3)=0f(1)=f(2)=f(3)=0 is impossible. That is because D1(1)=f(2)f(1)=0D_1(1)=f(2)-f(1)=0 and D1(2)=f(3)f(2)=0D_1(2)=f(3)-f(2)=0, implying D2(1)=D1(2)D1(1)=0D_2(1)=D_1(2)-D_1(1)=0. But as f(x)f(x) is a second degree polynomial, D2(x)D_2(x) is 2!=22! = 2 times the leading coefficient of f(x), so that leading coefficient is 00. But a polynomial can't have a leading coefficient of 00, so the assumption that f(1)=f(2)=f(3)f(1)=f(2)=f(3) was false.

I could also have proved this by using the fact that a kk-th degree polynomial has exactly kk (complex) roots, but it was nice that the Method of Differences could prove this as well.

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen Great analysis.

At this point, you should also question "Why did I need to deal with all of this, when Ang didn't?"

Hint: a1a=±1 a \mid 1 \Rightarrow a = \pm 1 .

Calvin Lin Staff - 6 years, 3 months ago

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I do not understand this notation. What do you mean by that?

Carlos E. C. do Nascimento - 6 years, 3 months ago

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I believe he is using the "divides" notation. That is, ab    kZ:ak=b a|b \iff \exists k \in \mathbb{Z} : ak = b

Andrew Edwards - 6 years, 3 months ago

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For someone not knowing that notation, your notation isn't really going to help, I'm afraid… ;-)

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen If aba|b then aa divides bb or aa is a factor of bb.

Victor Phan - 6 years, 3 months ago

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@Victor Phan Yeah, I'm aware of that. :)

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen Sorry! I was meant to say that to Carlos :D

Victor Phan - 6 years, 3 months ago

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We would love more awesome articles such as this..

Arshdeep Duggal - 6 years, 3 months ago

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