×

# Application of Method of Differences

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

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

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

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

Note by Calvin Lin
4 years, 1 month ago

Sort by:

Let $$g(n)=f(n+1)-f(n), h(n)=g(n+1)-g(n)$$. 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, $$f(2)|g(2)$$ and $$f(3)|g(3)$$. Hence $$f(2)|g(3)$$, so $$f(2)|h(2)$$. But $$h(1)=h(2)$$, so $$f(2)|h(1)=g(2)-g(1)$$, so $$f(2)|g(1)=f(2)-f(1)$$, so $$f(2)|f(1)$$.

Since $$f(1)$$ and $$f(2)$$ divide each other, they must have the same absolute value, which was what we wanted. · 4 years, 1 month ago

That's a great solution. I tried expressing $$f(2)$$ as some factor times $$f(1)$$ and $$f(3)$$ as some factor times $$f(2)$$, but I just got a big mess. · 4 years, 1 month ago

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. Staff · 4 years, 1 month ago

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

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

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

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

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

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

The right hand side is an integer, so the left hand side is too: this implies $$\frac{1}{a}$$ is an integer as well, so $$a\in \{ -1,1 \}$$. · 4 years, 1 month ago

This looks good.

However, why do you say that "the conditions imply $$a, f(1) \neq 0$$"? How do the conditions imply that? Why can't either of them be 0? Staff · 4 years, 1 month ago

Well, $$f(1) \mid f(2)$$, and $$0$$ doesn't divide anything, so $$f(1) \not = 0$$. Similarly, $$f(2) \mid f(3) \implies f(2) \not = 0$$. And as $$a=\frac{f(2)}{f(1)}$$, $$a$$ is not $$0$$ either. · 4 years, 1 month ago

Rethink that statement. Review the definition of $$|$$, which was given below.

In particular, Does $$0 \mid k$$ or $$k \mid 0$$ ? Staff · 4 years, 1 month ago

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

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

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

I could also have proved this by using the fact that a $$k$$-th degree polynomial has exactly $$k$$ (complex) roots, but it was nice that the Method of Differences could prove this as well. · 4 years, 1 month ago

Great analysis.

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

Hint: $$a \mid 1 \Rightarrow a = \pm 1$$. Staff · 4 years, 1 month ago

We would love more awesome articles such as this.. · 4 years, 1 month ago

I do not understand this notation. What do you mean by that? · 4 years, 1 month ago

I believe he is using the "divides" notation. That is, $$a|b \iff \exists k \in \mathbb{Z} : ak = b$$ · 4 years, 1 month ago

For someone not knowing that notation, your notation isn't really going to help, I'm afraid… ;-) · 4 years, 1 month ago

If $$a|b$$ then $$a$$ divides $$b$$ or $$a$$ is a factor of $$b$$. · 4 years, 1 month ago

Yeah, I'm aware of that. :) · 4 years, 1 month ago