Hello friends, there is a sum that tells, Find the integral values of \(n\), such that the expression

\(\large{\frac{n^2+n+3}{n+1}}\) is an integer.

When it is solved as a Diophantine equation, The answer comes out to be \( n= 0, 2\).

But checking it by putting the value \(n=-4\) it also satisfies.

Please explain.

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## Comments

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TopNewestNote that \(\frac{n^2+n+3}{n+1}\) is the same as \(n+\frac{3}{n+1}\).

So, for the expression to be integral; 3 should be divisible by \(n+1\).

And since we want only integral values of \(n\); they are \(2,0,-2,-4\) corresponding to the 4 factors \(3,1,-1,-3\).

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