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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.

Note by Md Zuhair
2 months, 3 weeks ago

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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 $$3,1,-1,-3$$. · 2 months, 3 weeks ago