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I've just spent past 5 days trying to get a solution to this:

Find all whole numbers, for whom the fraction

$\frac{n^3 + 2010}{n^2 + 2010}$

yields another whole number.

Before asking, please note that this is indeed an exercise from a math competition. However, this competition is already over, so don't worry; you're not doing my homework.

If anyone would be so kind to provide an explanation, could you formulate it in terms of high-school mathematics? Thanks.

P.S.: Please pardon my crude English. It's not my primary language.

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I have changed the title. Please use more descriptive titles. Also please avoid using words like "difficult" etc as they are subjective. –  Aryabhata Jan 16 '11 at 20:48

4 Answers 4

up vote 8 down vote accepted

Note $\rm\quad\displaystyle\ \frac{n^3+2010}{n^2+2010}\ =\ n + \frac{2010\ (1-n)}{n^2+2010}\ \in \ \mathbb Z\ \iff\ n^2+2010\ |\ 2010\ (1-n) $

$\rm\ gcd(n^2+2010,\: 1-n) = gcd(2011,\: 1-n)\:.\ $ Since $\:2011\:$ is prime, this $\rm gcd$ is $\:1\:$ if $\rm\ 1 < n < 2012\:,\ $ so if $\rm\ n < 2012\ $ then either $\rm\ n = 1\ $ or $\rm\ n^2+2010\ |\ 2010\ \Rightarrow \ n = 0\:.$

But $\rm\ \ n \ge 2012\ \Rightarrow\ n^2+2010\ > |2010\ (1-n)|\:.$

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That's the best answer up till now. The primality of 2011 is indeed the key to the solution. You forgot the case $n=1$ though. –  Raskolnikov Jan 16 '11 at 23:03
<strike>I don't understand the last implication. The fact that $n^2 + 2010 > |2010(1-2)|$ does not mean that $n^2+2010 \nmid 2010(1-n)$. Or does it?</strike> Aw crap, I get it now. –  MrZeta Jan 16 '11 at 23:45
All right, so this explains how to find out solutions for $n \in \mathbb{Z+}$, but (as I found out numerically and as WA confirmed) the fraction has another solution in $n = -2010$. Any idea how to find that one? –  MrZeta Jan 16 '11 at 23:54
@Raskolnikov: Originally it was a hint - hence the omitted minor details. I've since expanded it a bit. –  Bill Dubuque Jan 16 '11 at 23:55
@MrZeta: To handle $\rm\: n < 0\:$ put $\rm\:n\to-n\:$ above. –  Bill Dubuque Jan 17 '11 at 0:09

The idea to start with polynomial division is correct, but Raskolnikov is right concerning the end of the argument: the answer is a integer as long as the $2010 \frac{1-n}{n^2+2010}$ is an integer, say equal to $k$; you can then state that this is true whenever $\frac{n-1}{n^2+2010} = \frac{k}{2010}$ for $k \in \mathbb{Z}.$ Now if the function on the left becomes smaller than $1/2010,$ you won't have any more solutions; I hope this suffices to work out the rest of the problem.

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Could you please explain why is that? Isn't Z defined as {0, +-1, +-2, ...}? –  MrZeta Jan 16 '11 at 22:16
@MrZeta: I wasn't sure if you meant $\mathbb{N} = \{1, 2, 3, \cdots\}$ or $\mathbb{Z}$ (which is what you suggest, and what I call the set of integers). Exactly what point don't you understand? (By the way, I did cheat and change the sign of $k$ halfway through, sorry for that.) To me it seems easiest to try to find any integer solutions, otherwise you have to start worrying about signs. In the end you can discard any negative solutions. –  Gerben Jan 16 '11 at 22:26
I don't understand why the solution only exists when $f(x) >= \frac{1}{2010}$. I thought OK, that would make sense to me if $n \in \mathbb{N}$, but $n \in \mathbb{Z}$. –  MrZeta Jan 16 '11 at 22:31
@MrZeta: you're definitely right; I should have specified that I meant larger 'in absolute sense', $|f(n)|.$ –  Gerben Jan 16 '11 at 22:38
All right, now I feel I need only a little push to finish this. I've calculated the inequation, and found out that $-2010 <= n <= 0$ and $3 <= n <= 2007$. I suppose this could be right (got same result as @Myself did using gcd). I went desperate, punched the remainder into WA and was told that only integer results are 0, 1 and -2010. –  MrZeta Jan 16 '11 at 22:58

You find that (Euclidian algorithm)

$$\frac{ n^3 + 2010 }{ n^2 + 2010 } = n + \frac{ (1-n) 2010 }{ n^2 + 2010 }.$$

For a whole number division you need

$$\frac{ (1-n) 2010 }{ n^2 + 2010 } = k, $$

for some integer k. You can solve this last equation for 'n'

$$n = -\frac{1005}{k} \pm \frac{1}{k} \sqrt{1005} \sqrt{1005 - 2 k ( k - 1)}.$$

Now, noting that 1005 is not a square of an integer, you automatically conclude that

$$1005 - 2 k ( k - 1) = 1005 a^2,$$

for some integer 'a'. Solving this for 'k' yields

$$k = \frac{1}{2} \pm \frac{1}{2} \sqrt{2011 - 2010 a^2}.$$

Thus, the only solutions are those generated by a = 1, which are

$$k = 0 \Rightarrow n = 1;$$

$$k = 1 \Rightarrow n = 0 \text{ (positive sign solution) and } n = -2010 \text{ (minus sign solution).}$$

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We need to have $n^2+2010$ divide $n^3+2010$ evenly (without a remainder). After some polynomial division you find that

$$ n^3+2010 = n(n^2+2010)+2010-2010n $$

where the $2010-2010n$ is the remainder term. Thus the answer is a whole number only when $2010-2010n=0$, or $n=1$.

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Bugger. How could I forgot about polynomial division. But thanks very much. Aside from helping, you've showed me that I have something to revise :) –  MrZeta Jan 16 '11 at 20:42
I'm afraid the reasoning is not valid. I can give a counterexample to your method if instead of $2010$ we put the number $8$. Your method would give us the same result, but there is actually a solution for $n=4$. The reason is that while the rest term as a polynomial is not divisible by $n^2+2010$, it might well be divisible as a number. –  Raskolnikov Jan 16 '11 at 21:09

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