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Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer. I have no idea how to approach these types of proofs.

But I tried a few things, did not get me anywhere.

$n^3 -3 = an^2-7a$ then $n^3-an^2 = 3-7a$, and hence $n^2(n-a) = 3-7a$

And then I have no where to go...

Any help is appreciated thanks.

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3 Answers 3

up vote 14 down vote accepted

If $d|(n^3-3)$ and $d|(n^2-7)\implies d|\{n(n^2-7)-(n^3-3)\}\implies d|(3-7n)$

Again as $d|(n^2-7)$ and $d|(3-7n)\implies d|\{7(n^2-7)+n(3-7n)\}\implies d|(3n-49)$

Again as $d|(3n-49)$ and $d|(3-7n)\implies d|\{-3(3-7n)-7(3n-49)\}\implies d|334$

The rest is what precisely "Math Gems" has done.

So, $(n^2-7)|334$

Now, the factor$(F)$s of $334=2\cdot167$ are $\pm(1,2,167,334)$

Now, test for the integral solution of $n^2-7=F$

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2  
This is very sloppy! Firstly, and obviously, $(7n-3)/(n^2-7)$ can be an integer even if $n^2-7 > 7n-3$, as long as $7n-3$ is negative. Try $n=-3$, for instance. Secondly, your inequality $n > \frac{7+\sqrt{65}}{2} > 7$ is pure hand-waving: it is also true, for instance that $n > \frac{7+\sqrt{65}}{2} > 5$, but that doesn't mean we only have to test for $0 \le n \le 5$. For this to work, you must explicitly state that $7$ is the largest integer that is $\le \frac{7+\sqrt{65}}{2}$. –  TonyK Feb 20 '13 at 20:09
2  
@fosho: Be aware that this answer is wrong. $n=-3$ is a valid solution. –  TonyK Feb 20 '13 at 20:33
    
@TonyK, can you please have look into the rectified answer? –  lab bhattacharjee Feb 21 '13 at 6:09
1  
Isn't that just what Math Gems said, only more succinctly? –  TonyK Feb 21 '13 at 17:29
    
@TonyK, the idea of common divisor is totally different,right? Also I incorporated the acknowledgment in the answer. –  lab bhattacharjee Feb 22 '13 at 4:15

Hint $\rm\ \ n^2\!-\color{#0A0}7\mid n^3\!-\color{#C00}3\ \Rightarrow\ mod\,\ n^2\!-7\!:\,\ \color{#0A0}7^3\! = n^6 = \color{#C00}3^2\:\Rightarrow\: n^2\!-7\mid 7^3\!-3^2 =\, 2\cdot 167\ $

$\rm\,2,\,167\,$ are prime so, by unique factorization, $\rm\:n^2\!-7\mid\, 2\cdot 167\,$ $\Rightarrow$ $\rm\, n^2\!-7 = \pm\{1, 2,167, 334\},\,$ so $\ldots$

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Note that $$ \frac{n^3-3}{n^2-7} = n + \frac{7n-3}{n^2-7}. $$ So it suffices to determine when $n^2-7$ divides $7n-3$. It is not hard to show that $|n^2-7| > |7n-3|$ when $|n|>8$; so for those $n$, it is impossible for $n^2-7$ to divide $7n-3$. That leaves only the cases $n=-8,-7,\dots,8$ to check, revealing the two solutions $n=-3$ and $n=3$.

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