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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.2011-01-16

4 Answers 4

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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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    @MrZeta: Note above I used $\rm\: gcd(f(n),n-1) = gcd(f(1),n-1)\ $ for all polynomials $\rm\ f(n) \in \mathbb Z[n]$ by the polynomial **remainder theorem** (in case that step was not clear).2011-01-17
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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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    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 -2010.2011-01-16
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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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    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.2011-01-16