500 Followers Problem!

Observe that $y$ is even. let $y=2b$ $125b^2=27x^3+125$ It follows that $125 \mid 27x^3$ and $5 \mid x$ , so let $x=5a$ equation becomes $b^2=27a^3+1=(3a+1)(9a^2-3a+1)$ Now, let $d$ be the greatest common divisor of two factors in the RHS of this equation. We have $d \mid 9a^2-3a+1-3a(3a+1)=-6a+1 \ \ \ \ \Longrightarrow \ \ \ \ d \mid -6a+1+2(3a+1)=3 \ \ \ \ \Longrightarrow \ \ \ \ d \mid 3$ Therefore, $d=1$ or $d=3$ , but $3a+1 \equiv 9a^2-3a+1 \equiv 1 \pmod{3}$ , so $d=1$ . Note that product of two relatively prime factors is a perfect square. Hence both of them must be squares and specially $9a^2-3a+1=c^2$ Discriminant of this quadratic in terms of $a$ must be a perfect square too, this means $9-36(1-c^2)=k^2 \ \ \ \ \Longrightarrow \ \ \ \ (6c-k)(6c+k)=27$ One can easily find out that $c=\pm 1$ and $9a^2-3a+1=c^2=1 \ \ \ \ \Longrightarrow \ \ \ \ 9a^2-3a=0 \ \ \ \ \Longrightarrow \ \ \ \ a=0$ Which gives $x=0$ and $125y^2=500$ or $y^2=4$ . Thus the solutions are $(x, y)=(0, -2), (0, 2).$

Certainly $x$ must be a multiple of $5$ , and $y$ must be a multiple of $2$ . Writing $x = 5X$ and $y = 2Y$ we obtain that $Y^2 \; = \; 27X^3 + 1$ so that $a=3X\,,\,b=Y$ is a solution of the Diophantine equation $b^2 = a^3 + 1$ .

Euler proved that the only solutions in integers of this equation are $(0,1)\,,\,(0,-1)\,,\,(-1,0)\,,\,(2,3)\,,\,(2,-3)$ . Since we need $a$ to be a multiple of $3$ , we see that only the first two of these are relevant. Thus the only solutions to the original equation are $x=0\,,\,y = \pm2$ , and so the answer is $\boxed{2}$ .