Divisibility Test!

Algebra Level 4

Determine the sum of all positive integer $n$ such that $n^2 + 2$ divides $2 + 2001n$ .

The answer is 2016.

1 solution

Rimson Junio
Sep 13, 2015

If $(n^2+2)|(2+2001n)$ then $(n^2+2)|[(2+2001n)^2]$ or $(n^2+2)|(4+8004n+4004001n^2)$ . $\frac{4004001n^2+8004n+4}{n^2+2}=4004001+\frac{8004n-8007998}{n^2+2}$ This means that $(n^2+2)|(8004n-8007998)$ . We also note that $(n^2+2)|[8004n-8007998-4(2+2001n)]$ from which it follows that $(n^2+2)|8008006$ . It is easy to check all the factors of 8008006 and that the solution set for $n$ is {6,9,2001}.

You are awesome, dude!

tanay gaurav - 5 years, 9 months ago

In your first statement, if $(n^2+2)$ divides $(2+2001n)$ , why does $(n^2+2)$ divide $[(2+2001n)^2)]$ ?

Deb Sen - 5 years, 9 months ago

Suppose $2 + 2001n = k(n^{2} + 2)$ where k is an integer. Hence $(2 + 2001n)^{2} = k_{1}(n^{2} + 2)$ where $k_{1} = k^{2}(n^{2} + 2)$ is an integer (since n and k are integers).

Krutarth Patel - 5 years, 9 months ago

Whoops, obvious enough, thanks for the clarification though!

Deb Sen - 5 years, 8 months ago

Great approach!

Karan Arora - 5 years, 9 months ago

Sir in step which you wrote (n^2+2)|{8004n-8007998-4(2+2001)} where did that {-4(2+2001)} term come?

avn bha - 5 years, 7 months ago

Since $n^{2} + 2 | (8004n - 8007998)$ , let $8004n - 8007998 = k_{1}(n^{2} + 2)$ and $n^{2} + 2 | 2 + 2001n$ , let $2 + 2001n = k_{2}(n^{2} + 2)$ . Now, $8004n - 8007998 - 4(2 + 2001n) = (k_{1} - 4k_{2})(n^{2} + 2)$ , hence $n^{2} + 2 | (8004n - 8007998 - 4(2 + 2001n))$ . This can be extended to show that for integers $a$ , $b$ and $c$ such that $c | b$ and $c | a$ , $c | (xa + yb)$ for $x, y \in \mathbb{Z}$ . Also, the term was introduced to get rid of the term containing $n$ .

Krutarth Patel - 5 years, 7 months ago

thanks ! i got it

avn bha - 5 years, 7 months ago

I didnt get the 8007998

Mr Yovan - 5 years, 4 months ago

Divisibility Test!

The answer is 2016.

1 solution

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