The best way to do this is say: If n is an integer and such an expression is also, we can say that their product is an integer, 4n2/(n2 +3) is an integer. Now we can add 12 and then substract it so we get 4n²/(n² + 3) = 4 + 12/(n² +3) which is an integer if and only if (n² +3)|12 etc. n {0, 1, -1, 3, -3} We must go back to the original expression to make sure that they work.
Trivially, n = 0 works Let's find all non-zero n such that |n^2 + 3| > |4n|, we can dispel these Note that for positive n, we can remove the absolute value symbol Solving for: n^2 - 4n + 3 > 0 n^2 - 4n + 3 = (n - 1)(n - 3) > 0 This happens when n > 3 or n < 1 Thus, assuming n > 0, we only need to check when n is 1, 2, or 3 Now, note that if n is positive, and m = -n, then |4n| = |4m| and |n^2 + 3| = |m^2 + 3|, thus, among negative n, we only need to check -1, -2, -3 Furthermore, it is easy to see that if a positive n works, its negative counterpart -n also works, so we'll only check 1, 2, and 3 n = 1 works because we get 4/4 = 1 n = 2 does not work because we get 8/7 n = 3 works because we get 12/12 = 1 Thus, the answers are 0, 1, 3, -1, -3
Due to respect, how about we can just let a be an integer a= 4n/(n^2 +3) then you will get an^2 -4n + 3a = 0 by arranging (an -3)(n-a), n=a so n is also integer n= -3/a then a can be 1,-1,3,-3 which are divisible for -3 and are the solutions of n. You might ask what happens you arrange like (an-3/2)(n-2a) etc.... The out-come is same if you simply it. Math is a language. Communicate simply.
The best way to do this is say: If n is an integer and such an expression is also, we can say that their product is an integer, 4n2/(n2 +3) is an integer. Now we can add 12 and then substract it so we get 4n²/(n² + 3) = 4 + 12/(n² +3) which is an integer if and only if (n² +3)|12 etc. n {0, 1, -1, 3, -3} We must go back to the original expression to make sure that they work.
Trivially, n = 0 works
Let's find all non-zero n such that |n^2 + 3| > |4n|, we can dispel these
Note that for positive n, we can remove the absolute value symbol
Solving for: n^2 - 4n + 3 > 0
n^2 - 4n + 3 = (n - 1)(n - 3) > 0
This happens when n > 3 or n < 1
Thus, assuming n > 0, we only need to check when n is 1, 2, or 3
Now, note that if n is positive, and m = -n, then |4n| = |4m| and |n^2 + 3| = |m^2 + 3|, thus, among negative n, we only need to check -1, -2, -3
Furthermore, it is easy to see that if a positive n works, its negative counterpart -n also works, so we'll only check 1, 2, and 3
n = 1 works because we get 4/4 = 1
n = 2 does not work because we get 8/7
n = 3 works because we get 12/12 = 1
Thus, the answers are 0, 1, 3, -1, -3
Interesting math problem!
Due to respect, how about we can just let a be an integer a= 4n/(n^2 +3)
then you will get an^2 -4n + 3a = 0 by arranging
(an -3)(n-a), n=a so n is also integer n= -3/a then a can be 1,-1,3,-3 which are divisible for -3 and are the solutions of n.
You might ask what happens you arrange like (an-3/2)(n-2a) etc....
The out-come is same if you simply it.
Math is a language. Communicate simply.