What about This? Please help

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.

#NumberTheory

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Comments

Note 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 $3,1,-1,-3$ .

Yatin Khanna - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$