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From Fundamentals of Number Theory by LeVeque, section 3.1, prob. 1

Let $f(x) = a_0x^n + \cdots + a_n$ be a polynomial over Z. Show that if $r$ consecutive values of $f$ (i.e., values for consecutive integers) are all divisible by $r$, then $r|f(m)$ for all $m \in Z$. Show by an example that this can happen with $r \gt 1$, even when $(a_0, \cdots, a_n) = 1$.

Can someone give me a hint on how to proceed? (not a full answer please.. just a nudge in the right direction.) I tried playing around with the division algorithm and looking for $r$ as as the $GCD(f(m), f(x))$ but I'm not making any headway. Seems to me the only way this can happen is if $r|a_i$ across the board but the last line about 'even when $(a_0, \cdots, a_n) = 1$' says that's not true..

Thanks in advance.

Also props to anyone who can think of a specific applications of this property in a non-number theory situation like an optimization or how this might come up when coding.


Possible Answer -- Thanks for all the help guys! Here's what I came up with after fiddling with your suggestions:

Starting with the binomial theorm:

$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$

I write,

$(rp + q)^n = \sum_{k=0}^n {n \choose k}(rp)^{k}q^{n-k}= {n \choose 0}q^n$ mod $r= q^n$ mod $r$

$\Rightarrow f(rp+q) \equiv f(q)$ mod $r$

Now, since (1) any $m \in Z$ can be written as $m=rp+q$ for some $p,q \in Z$, and (2) we know the $r$ consecutive values of integers cover all members of $Z_r$, and (3) $r|f(q)$ for all $q \in Z_r$ then by (1), (2), (3) and the binomial calculation above $r|f(m)$ for all $m \in Z$.

I think that works. Can anyone confirm?

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If $x \equiv y (\mod r)$, what can you saw about $f(x)$ and $f(y)$ modulo $r$? –  John M Jul 17 '11 at 19:30
    
For the example, look at $x^2 - x$; it shouldn’t be hard to find an appropriate $r$. –  Brian M. Scott Jul 17 '11 at 19:33
    
@unclejamil: Looks good. But simply the fact that exponentiation (or multiplication, really) is well-defined in the residue class ring may save a bit of ink: $x\equiv y \pmod r$ $\Rightarrow$ $x^n\equiv y^n \pmod r$ for all natural numbers $n$ $\Rightarrow$ $f(x)\equiv f(y)\pmod r.$ IOW while going via the binomial formula works all right, you really only need: $a\equiv b, c\equiv d \Rightarrow ac\equiv bd$ (all modulo $r$). –  Jyrki Lahtonen Jul 18 '11 at 4:42
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3 Answers 3

up vote 2 down vote accepted

OK, a small hint to start with, maybe expandable. Let $r=2$ and $f(x)=x(x-1)$.

And another hint: Take $r$ consecutive integers, and look at their remainders on division by $r$. What do you get?

And another hint: Corollary to Theorem 3.2.

Added: The Binomial Theorem proof that was added to your post is correct, well written, and shows initiative.

Here is an alternative version that would have been standard at the time of Euler, before Gauss introduced the congruence relation. The main diffference is that Euler would have written $a$, $b$, $\dots$, $z$ instead of using subscripts!

Lemma: Let $f(x)=a_0x^n +a_1x^{n-1}+\cdots + a_n$ be a polynomial with integer coefficients. Suppose that $r$ divides $x-y$. Then $r$ divides $f(x)-f(y)$.

Proof: We have $$f(x)-f(y)=a_0(x^n-y^n)+a_1(x^{n-1}-y^{n-1}) +\cdots+a_{n-2}(x^2-y^2)+a_{n-1}(x-y).$$

If we can show that $x-y$ divides $x^k-y^k$ for all $k$, we will be finished. But note that $$x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+ \cdots +xy^{k-2}y+y^k).$$

The identity that we used is not mysterious. It is a slightly fancied up version of the familiar $$1-x^k=(1-x)(1+x+x^2+\cdots +x^{k-1})$$ or (if $x \ne 1$) equivalently $$1+x+x^2+\cdots+x^{k-1}=\frac{1-x^k}{1-x}$$ which we use in summing a finite geometric series.

The reason I point it out is that it is nice to make connections between various parts of mathematics. That said, it is very important that you master the modern algebraic point of view. LeVeque takes such a point of view, though, given his era, it is less uncompromisingly algebraic than more modern serious books.

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This was very helpful. Thanks for the pointers and the perspective. –  unclejamil Jul 19 '11 at 19:42
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Others have given you good hints, so I will add a few bits related to your last question. The problem of identifying the set of polynomials with integer coefficients whose values all vanish modulo a given modulus $m$ has obviously been studied and solved a long time ago, but the solution is not too well known except in the case $m=p$, a prime, when it naturally pops up in the context of Little Fermat.

Recently I had a reason to dig out this piece of elementary number theory. A solution has been published by Singmaster in 1974 and Niven & Warren in 1957. The earliest hit that I found was in a paper by Carlitz (pre WWII), who says that this was an exercise in a classical algebra textbook by Dickson. I don't claim this list to be comprehensive by any means, but it does show that this particular problem has been forgotten and rediscovered many times over.

An application to coding? Well, here's one - I'm running the risk of blowing my own trumpet here, sorry about that.

A bit of background. It is easy to see that a quadratic polynomial $ax+bx^2$ with integer coefficients gives rise to a permutation of $\mathbf{Z}/m\mathbf{Z}$, if (it is essentially also only if, but let's skip that part) $a$ is coprime to $m$ and $b$ is nilpotent mod $m$ (i.e. divisible by all the prime factors of $m$). Takeshita and Sun (working at Ohio State EE department at the time) noticed that this kind of permutations can be tuned up to work as bit interleavers in an error-correcting coding scheme known as turbo coding. A set of such polynomials (for a sequence of values of $m$) was adopted by an industrial standardization body of all the major cellular players (operators, cellphone and base station HW manufacturer etc) to a 4G cellular standard. I was working for a largish Finnish cell phone manufacturer at the time, and my colleagues had a need to compute the inverse function. At that time we looked at good enough ad hoc -solutions, but later I returned to the problem of finding the least degree polynomial inverse function. We ended up using the binomial series for the solution (the usual formula for the quadratic equation), because the condition on $b$ tells that the series truncates to a polynomial, when we view the coefficients modulo $m$. Then we can reduce the degree of that truncated series, because Niven&Warren and Singmaster give a Gröbner basis of the ideal of vanishing polynomials.

Not one of the hottest pieces of math, but, hey, it earned me a trip to Dublin, a guided tour to Jameson distillery and such!

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Very nice example! I went and read up on turbo coding and eventually ended up at spectrum.ieee.org/computing/software/… -- super interesting. I've got your paper on my to-do list as well. :) Thanks for the help. BTW: I sent this on to a friend at work who had just asked me about how this (number theory) was useful. He was really impressed with this application. –  unclejamil Jul 19 '11 at 19:46
    
@unclejamil: Glad to hear that you find this interesting. Actually I'm supposed to be revising our manuscript this very minute (we got relatively positive reviews from IEEE Transactions on Information Theory). But MathSE is just so much more fun... –  Jyrki Lahtonen Jul 19 '11 at 19:59
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Hint $\ $ Every sequence of $\rm\:r\:$ consecutive integers includes all possible values modulo $\rm\:r,\:$ therefore for some $\rm\:b_i\:$ in the given sequence $\rm\:m\equiv b_i,\:$ thus $\rm\:f(m)\equiv f(b_i)\equiv 0\ \ (mod\ r).\: $ Recall that congruences are preserved by addition and multiplication,$\:$ so they are also preserved by compositions of such. $ $ In particular, congruences are preserved by applying a polynomial $\rm\:f(x)\:$ with integer coefficients.

For example, for $\rm\:f(x) = x^2 + a\ x + b,\:$ $\rm\:2\ |\ f(0), f(1)\:$ $\Rightarrow$ $\rm\:b\equiv 0,\ a\equiv 1\:$ $\Rightarrow$ $\rm\,2\:|\:x\,(x+1),\ \forall\, x.$

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