200 Followers Problem!

Let $A=200^2-200$ . Thus $r^3-s^3=Ars$ . Write $r-s=n$ , where $n \in \mathbb{Z}$ . Then $(s+n)^3-s^3=A(s+n)s$ , or $(A-3n)s^2+(An-3n^2)s-n^3=0$ According to quadratic formula, we get $s=-\frac{n}{2}\pm \frac{|n|}{2} \frac{\sqrt{(A+n)(A-3n)}}{A-3n}$ Therefore, $(A+n)(A-3n)=k^2$ , where $k$ is a non-negative integer. It follows that $-A \le n \le \lfloor A/3 \rfloor$ .

Note that $A$ is divisible by $200$ . In the equation $(A+n)(A-3n)=k^2$ look mod $5$ , then $-3n^2 \equiv k^2 \pmod{5}$ . Hence $n \equiv k \equiv 0 \pmod{5}$ , since a perfect square mod $5$ is congruent to $0, 1, 4$ only. If you divide both sides of the equation by $25$ and look mod $5$ again, then it comes out that $n \equiv k \equiv 0 \pmod{25}$ .

In the equation $(A+n)(A-3n)=k^2$ look mod $8$ , then $-3n^2 \equiv k^2 \pmod{8}$ . Hence $n$ and $k$ are both even, since a perfect square mod $8$ is congruent to $0, 1, 4$ only. If you divide both sides of the equation by $4$ and look mod $8$ again, then a similar argument gives $n \equiv k \equiv 0 \pmod{4}$ . Now write $n=4n'$ and $k=4k'$ and divide both sides of the equation by $16$ to get $(A/4+n')(A/4-3n')=k'^2$ . Look mod $8$ , then $4+4n'+5n'^2 \equiv k'^2 \pmod{8}$ . If $n'$ is odd, then LHS of this congruence is $5$ mod $8$ , which is contradiction. Thus $n'$ and $k'$ are both even and finally $n \equiv k \equiv 0 \pmod{8}$ .

Thus in total we have that $n \equiv k \equiv 0 \pmod{200}$ . Write $n=200 N$ and $k=200 K$ to get $(199+N)(199-3N)=K^2$ For $\color{#D61F06}N=-199$ and $\color{#D61F06}N=0$ the LHS becomes a square. Then, for $N \not\equiv 0 \pmod {199}$ we have that $d=\gcd(199+N, 199-3N) \mid 3(199+N)+199-3n=4\times 199$ Hence $d=1, 2, 4$ .

If $d=1$ , then $199+N=a^2$ and $199-3N=b^2$ , where $a$ and $b$ are co-prime odd positive integers. Hence $3a^2+b^2=796$ . Its easy to see that $a \le 15$ . Checking $a=1, 3, 5, ..., 15$ , reveals that $a=13, 15$ make $796-3a^2$ a perfect square. Thus we get two pairs: $\color{#3D99F6}(a, b)=(13, 17), (15, 11)$ that result in $\color{#D61F06}N=-30$ and $\color{#D61F06}N=26$ , respectively.

If $d=2$ , then $199+N=2a^2$ and $199-3N=2b^2$ , where $a$ and $b$ are co-prime positive integers. It follows that $3a^2+b^2=398$ . Its easy to see that $a \le 11$ . Check that none of $a \le 11$ make $398-3a^2$ a perfect square.

If $d=4$ , then $199+N=4a^2$ and $199-3N=4b^2$ , where $a$ and $b$ are co-prime positive integers. It follows that $3a^2+b^2=199$ . Its easy to see that $a \le 7$ . Checking $a \le 7$ , reveals that $a=1$ makes $199-3a^2$ a perfect square. Thus we get one pair: $\color{#3D99F6}(a, b)=(1, 14)$ that results in $\color{#D61F06}N=-198$ .

Finally we get $5$ solutions for $n=200N$ , namely $n=-39800, -39000, -6000, 0, 5200$ $n=-39800$ gives only $s=19900$ . $n=0$ gives $A=0$ , which is a contradiction! Other values of $n$ give two different solutions for $s$ . Thus the total number of solutions for the original equation is $\boxed{7}$ .

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Notice that if $(r, s)$ is a solution, then $(-s, -r)$ is a solution too. Thus I write four of the solutions to the original equation. $(r, s)=(s+n, s)=\left(-19900, 19900 \right), \left(-\frac{126750}{7}, \frac{146250}{7} \right), \left(-\frac{12000}{17}, \frac{90000}{17} \right), \left( \frac{67600}{11}, \frac{10400}{11} \right)$

The equation becomes $r^3 - s^3 \; =\; N(N-1)rs$ where $N=200$ , and we want $r = s+n$ for some integer $n$ . Thus $\begin{aligned} 3s^2n + 3sn^2 + n^3 & = N(N-1)(s^2 + sn) \\ \big[N(N-1) - 3n\big]s^2 + n\big[N(N-1) - 3n\big]s - n^3 & = 0 \end{aligned}$ and so $s \; = \; -\tfrac12n \pm \frac{n}{2\big[N(N-1) - 3n\big]}\sqrt{\big(N(N-1)-3n\big)\big(N(N-1) + n\big)}$ Thus $-N(N-1) \le n < \tfrac13N(N-1)$ , and $\big(N(N-1) - 3n\big)\big(N(N-1) + n\big)$ must be a perfect square.

Simple calculations with $N=200$ shows that there are only $5$ values of $n$ for which this is true: $-39800, -39000, -6000, 0, 5200$ .

When $n=-39800$ , the only value of $s$ is $19900$ . When $n=0$ , the only value of $s$ is $0$ , and this is not a permitted solution of the original equation. The other three values of $n$ each yield two values of $s$ , and so there are a total of $\boxed{7}$ solutions.