A number theory problem by Md Zuhair

Number Theory Level 3

n 2 + n + 3 n + 1 \large{\frac{n^2+n+3}{n+1}}

Find the sum of all the integral values of n n , such that the expression above is an integer.


The answer is -4.

Md Zuhair by Md Zuhair , 20, India

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1 solution

Achal Jain
Jan 28, 2017

We can break the equation as

n ( n + 1 ) + 3 n + 1 = n ( n + 1 ) n + 1 + 3 n + 1 = n + 3 n + 1 \frac { n(n+1)+3 }{ n+1 } =\frac { n(n+1) }{ n+1 } +\frac { 3 }{ n+1 } =\quad n+\frac { 3 }{ n+1 } .

Now we want it to be an integer therefore n + 1 ∣ 3 n+1|\quad 3 .

The possible values are − 4 , − 2 , 2 , 0 -4,-2,2,0 .

Hence the sum is − 4 + − 2 + 2 + 0 = − 4 -4+-2+2+0=-4

3 Helpful 0 Interesting 0 Brilliant 0 Confused

What if I told you missed n=0? (You used wrong condition!)

Rishabh Jain - 4 years, 4 months ago

@Rishabh Cool what is wrong with the condition?

Achal Jain - 4 years, 4 months ago

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Take n=5, the condition you mentioned in your solution is satisfied but n=5 isn't correct, the point is why are you assuming its reciprocal to be an integer too if a fraction is an jnteger?

Rishabh Jain - 4 years, 4 months ago

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