+0  
 
0
1
1
avatar+24 

What is the sum of all positive integers "q" such that (n^7 - n^3)/q is an integer for every positive integer n < 2021?

 

Thanks in advance!

 May 21, 2024
 #1
avatar+621 
-1

We can analyze why the expression (n^7 - n^3)/q must be an integer for every positive integer n less than 2021.

 

Factoring: First, we can factor the expression: (n^7 - n^3) = n^3 (n^4 - 1). We can further factor (n^4 - 1) using the difference of squares pattern: (n^4 - 1) = (n^2 + 1)(n^2 - 1).

 

Divisibility by n^2 + 1 and n^2 - 1: For the expression (n^7 - n^3)/q to be an integer for every positive integer n less than 2021, either n^2 + 1 or n^2 - 1 (or both) must be divisible by q.

 

Here's why:

 

If n is even, then n^2 is even, making (n^2 + 1) odd and not divisible by q. In this case, (n^2 - 1) must be divisible by q for the entire expression to be an integer.

 

If n is odd, then n^2 is odd, making (n^2 - 1) even and not divisible by q. In this case, (n^2 + 1) must be divisible by q for the entire expression to be an integer.

 

Restriction on q: Since q must divide both (n^2 + 1) and (n^2 - 1) for all positive integers n less than 2021, it must be a factor of their greatest common divisor (GCD). The GCD of (n^2 + 1) and (n^2 - 1) is always 2, as for any positive integer n, one of these terms will be even and the other odd.

 

Therefore, the only possible value for q is 2.

 

In conclusion, the sum of all positive integers q that satisfy the condition is simply the value of q itself: 2. So the answer is 2.

 May 21, 2024

5 Online Users

avatar
avatar
avatar
avatar
avatar