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Find the number of values of $N$ such that the below expression is an integer: $(n+1)^2\over n+7$ is an integer

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up vote 5 down vote accepted

$(n+1)^2=n^2+2n+1=(n+7)(n-5)+36$

So, $\frac{(n+1)^2}{n+7}=n-5+\frac{36}{n+7}$

Assuming $n$ to be an integer, $(n+7)\mid36 \iff (n+7)\mid(n+1)^2$

So, $n+7$ can be any divisor of $36,$ namely $\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$

If we constrain $n$ to be non-negative i.e., if $n+7\ge 7,$ then $n+7$ can be $9,12,18,36$

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I don't think this is right; for example $n=65$ satisfies your criterion, but not the equation required by the asker. –  Fixee Nov 16 '12 at 15:04
    
@RossMillikan, thanks Ross, you took my burden. –  lab bhattacharjee Nov 16 '12 at 15:10
    
@RossMillikan: He edited it: he had the condition "$(n+7)\ |\ 36m$ for integer $m$" as a condition. –  Fixee Nov 16 '12 at 15:10
    
@Fixee, are you ok with the current version? Also, I don't know where the version you are talking about is lost. –  lab bhattacharjee Nov 16 '12 at 15:21
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