700 Followers Problem!

If we write $n = a+b$ then we have $\begin{aligned} n^3 - 3abn + 3(n^2 - 2ab) & = 700n^2 \\ ab & = \frac{n^2(n-697)}{3(n+2)} \end{aligned}$ so that $a,b$ are the roots of the quadratic equation $X^2 - nX + \frac{n^2(n-697)}{3(n+2)} \; = \; 0$ Thus the discriminant of this quadratic must be the square of a rational, and hence $\frac{n^2(2794 - n)}{3(n+2)} \; = \; q^2$ for some rational $q$ . This means that $-2 < n \le 2794$ and that $\begin{aligned} (n+2)(2794 - n) & = 3k^2 \\ 5588 + 2792n - n^2 & = 3k^2 \\ (n - 1396)^2 + 3k^2 & = 1954404 = 2^2 \times 3^2 \times 233^2 \end{aligned}$ for some non-negative integer $k$ . Considering the factor of $3$ , it is clear that $n-1396$ , and hence also $k$ , is a multiple of $3$ . Thus $n = 1396 + 3u$ , $k = 3v$ where $u,v$ are integers with $v \ge 0$ and $u^2 + 3v^2 \; = \; 2^2 \times 233^2$ Since $233 \equiv 1 \pmod{4}$ we see that $\left(\frac{-3}{233}\right) \; = \; \left(\frac{-1}{233}\right) \left(\frac{3}{233}\right) \; = \; \left(\frac{233}{3}\right) \; = \; \left(\frac{2}{3}\right) \; = \; -1$ and so $-3$ is not a quadratic residue modulo $233$ . Thus $v$ , and so $u$ as well, is a multiple of $233$ . Thus we deduce that $n = 1396 + 699p$ , $k = 699q$ , where $p,q$ are integers with $q \ge 0$ and $p^2 + 3q^2 = 4$ . There are only four possible solutions, namely $(p,q) = (2,0), (-2,0), (1,1), (-1,1)$ , which yield $n = -2, 2794, 697, 2095$ respectively. Ignoring the first of these (since we know that $n > -2$ ), there are three possible values of $n$ , with one solution for $a,b$ for each such value, and so the answer to the question is $2794+697+2095 = \boxed{5586}$ .

Let $k=a+b$ and rewrite the equation: $k^3-3abk+3(k^2-2ab)=700k^2 \ \ \Longrightarrow \ \ ab=k^2 \frac{k-697}{3k+6} \ \ \Longrightarrow \ \ a^2-ka+k^2 \frac{k-697}{3k+6}=0 \qquad (1)$ This is a quadratic in $a$ and hence its discriminant must be a square of a rational. Thus $\Delta_a=k^2-4k^2 \frac{k-697}{3k+6}=k^2 \frac{2794-k}{3(k+2)} \qquad (2)$ is square of a rational and consequently $\frac{2794-k}{3(k+2)}$ is non-negative and square of a rational. Observe that $-1 \le k \le 2794$ . Check that for $k=-1$ the expression $\frac{2794-k}{3(k+2)}$ is not square of a rational. For $k=0$ and $k=2794$ , discriminant becomes zero and solving quadratic equation $(1)$ together with $k=a+b$ yields to two solutions from here: $\color{#D61F06} (a, b)=(0, 0), (1397, 1397)$ .

Henceforth, $0<k<2794$ and for co-prime positive integers $p, q$ $\frac{2794-k}{3(k+2)}= \frac{p^2}{q^2} \ \ \Longrightarrow \ \ k=\frac{2794 q^2-6p^2}{3p^2+q^2}$ Thus, $3p^2+q^2 \mid 2794 q^2-6p^2$ or $3p^2+q^2 \mid 2794 q^2-6p^2+2(3p^2+q^2)=2796 q^2 \qquad (3)$ Notice that $d=\gcd(q^2, 3p^2+q^2)=\gcd(q^2, -q^2+3p^2+q^2)=\gcd(q^2, 3p^2)=\gcd(q^2, 3)=1, 3$ And $d=3$ occurs when $3 \mid q$ . Suppose that $d=1$ . Then from relation $(3)$ we have $3p^2+q^2 \mid 2796=3 \times 4 \times 233$ , but $3 \nmid q$ , and this implies that $3p^2+q^2 \mid 932=4 \times 233$ . Hence possible values for $3p^2+q^2$ are $4, 233, 466$ and $932$ . We claim that $3p^2+q^2=4$ . We use the facts that a perfect square can only be congruent to $0, 1$ mod $4$ and congruent to $0, 1, 4, 7$ mod $9$ . Note that $3p^2+q^2$ mod $4$ never produce $2$ and, thus, $3p^2+q^2=466$ is not possible because $466 \stackrel{4}{\equiv} 2$ . Also, $3p^2+q^2$ mod $9$ never produce $5$ and $8$ . Hence $3p^2+q^2=233$ and $3p^2+q^2=932$ are not possible because $233 \stackrel{9}{\equiv} 8$ and $932 \stackrel{9}{\equiv} 5$ . So $3p^2+q^2=4$ and $p=q=1$ , which gives $k=697$ and equation $(1)$ gives another solution, namely $\color{#D61F06} (a, b)=(697, 0)$ .

Remaining case is $d=3$ , where $q=3q'$ for a positive integer $q'$ . Then relation $(3)$ gives $3p^2+9q'^2 \mid 2796 \times 9q^2.$ And since $3 \nmid p$ , we get $3p^2+q'^2 \mid 4 \times 233$ . By the same arguments above we get $p=q'=1$ . Therefore $p=1$ and $q=3$ .This yields to $k=2095$ . Using equation $(1)$ gives the last solution $\color{#D61F06} (a, b)=\left(4190/3, 2095/3 \right)$ .

Then the final answer to this problem is $\sum k= 0+2794+697+2095= \boxed{5586}$ .