But, That's a '+' sign!

Number Theory Level 2

Find the sum of all non-negative integers $n$ such that $n+1|n^2+1$ .

The answer is 1.

3 solutions

Nihar Mahajan
Jun 19, 2015

$n^2+1 = n^2-1+2 = (n+1)(n-1)+2 \\ \Rightarrow n+1 \ | \ (n+1)(n-1)+2 \\ \Rightarrow n+1 \ | \ 2$

So we have 2 cases:

Case 1: $n+1=1 \Rightarrow \boxed{n=0}$

Case 2: $n+1=2 \Rightarrow \boxed{n=1}$

So sum $=1+0 =\boxed{1}$ .

@Mehul Arora , yeah thats a " + " sign , but I was powerful enough to make it " - " sign :3 :3 :3 :P

Nihar Mahajan - 5 years, 12 months ago

Adarsh Kumar
Jun 19, 2015

Since $(n+1)|(n^2+1)\\ \it\ would\ also\ divide\\ n^2+1+2n+2 \\=(n+1)^2+2\\ \Longrightarrow (n+1)|2$ .Now it is easy but Mehul Arora you have to mention that $n$ is positive as i first answered $-4$ .

Moderator note:

For such problems on division by linear terms, $(n -a) \mid f(n) \Rightarrow (n - a) \mid f(a)$ .

Do you see why? What theorem does that look like?

And mention that it must be an integer

Mohamed Wafik - 5 years, 12 months ago

Yes!That too!

Adarsh Kumar - 5 years, 12 months ago

Nobita Nobi
Jun 26, 2015

Common sense

But, That's a '+' sign!

The answer is 1.

3 solutions

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