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 \( f(2) = \pm f(1) \).

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 \( f(2) = \pm f(1) \).

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TopNewestLet \(g(n)=f(n+1)-f(n), h(n)=g(n+1)-g(n)\). 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. – Ang Yan Sheng · 4 years, 1 month ago

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– Tim Vermeulen · 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.Log in to reply

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 · 4 years, 1 month ago

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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 \(a\in \{ -1,1 \}\). – Tim Vermeulen · 4 years, 1 month ago

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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? – Calvin Lin Staff · 4 years, 1 month ago

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– Tim Vermeulen · 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.Log in to reply

In particular, Does \( 0 \mid k \) or \( k \mid 0\) ? – Calvin Lin Staff · 4 years, 1 month ago

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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. – Tim Vermeulen · 4 years, 1 month ago

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

Hint:\( a \mid 1 \Rightarrow a = \pm 1 \). – Calvin Lin Staff · 4 years, 1 month agoLog in to reply

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

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

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– Andrew Edwards · 4 years, 1 month ago

I believe he is using the "divides" notation. That is, \( a|b \iff \exists k \in \mathbb{Z} : ak = b \)Log in to reply

– Tim Vermeulen · 4 years, 1 month ago

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

– Victor Phan · 4 years, 1 month ago

If \(a|b\) then \(a\) divides \(b\) or \(a\) is a factor of \(b\).Log in to reply

– Tim Vermeulen · 4 years, 1 month ago

Yeah, I'm aware of that. :)Log in to reply

– Victor Phan · 4 years, 1 month ago

Sorry! I was meant to say that to Carlos :DLog in to reply