A number theory problem by Akshat Sharda

Let $k = n - 1$ . Then $a = \dfrac{(k + 4)^{3}}{k} = \dfrac{k^{3} + 12k^{2} + 48k + 64}{k} = k^{2} + 12k + 48 + \dfrac{64}{k}$ .

So $a$ will be an integer as long as $k | 64$ , i.e., for $k = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64$ .

We thus have $m = 7*2 = 14$ , and since $n = k + 1$ , while all the $k$ 's add to zero, the $14$ $n$ 's will add to $14*1 = 14$ . So the desired sum is $14 + 14 = \boxed{28}$ .

n-1|64 therefore no. Of valyes of such n = divisors of 64. =7+7. Other 7 is for negative n. And sum of these values of n-1 is sum of positive and negative divisors of 64=0. Thefore sum of such n is 14. Hence answer=14*2

There is a theorem that affirmates that the rest of a polynom division P(n)/n-a is equal to P(a).Therefore,a=Q(n)+64/(n-1),where Q(n) is the quocient of (n+3)^3/(n-1) and 64=P(1).Thus, a belong to integers if,and only if,64/(n-1) is an integer,which means (n-1) is a divisor of 64.We know 64=2^6,so we have 14 solutions(7 positive and 7 negative).If we sort the solutions by a crescent order,the difference between a positive and a negative solution is equal 2.Then,The answer is 2.7+14=28