900 Followers Problem!

$\boxed{{\color{#D61F06}\text{First solution:}}}$

Let $\frac{m^3-n^3}{m^2+n^2-mn‎}=p,$ where $p$ is a prime number. Observe that $m>n$ . Let $d=\gcd(m, n)$ , then there are relatively prime numbers $a$ and $b$ such that $m=ad$ and $n=bd$ . We have $a>b$ and the equation becomes $d(a^3-b^3)=‎p(a^2+b^2-ab)‎‏ \ \ \ \Rightarrow \ \ \ d(a-b)‎\big(a^2+b^2+ab\big)‎=‎p\big( (a-b)^2+ab\big). \qquad (1)$ The LHS of equation $(1)$ is divisible by $a-b$ , and so is the RHS. But note that ‎ $\gcd\big(a-b, ‎(a-b)^2+ab\big)=\gcd\big(a-b, ‎-(a-b)^2+(a-b)^2+ab\big)=\gcd(a-b, ‎ab)‎.$ If $a-b$ and $ab$ are not relatively prime, there is a prime number $q$ such that $q|ab$ and $q|a-b$ . Hence $q$ divides one of $a$ or $b$ . For example, if $q|a$ , then $q|a-b$ forces $q|b$ , which contradicts $\gcd(a, b)=1$ . Hence $\gcd(a-b, ab)=1$ and consequently $a-b$ and $(a-b)^2+ab$ are relatively prime. Then, from equation $(1),$ we have $a-b|p$ , and, therefore, $a-b=1$ or $a-b=p$ . If $a-b=p$ equation $(1)$ becomes $d‎\big(a^2+b^2+ab\big)‎=‎a^2+b^2-ab$ , which is impossible (LHS is greater than RHS). Thus $a-b=1$ and‎ equation $(1)$ becomes $d‎\big((b+1)^2+b^2+b(b+1)\big)‎=‎p\big(1+b(b+1)\big) \ \ \ \Rightarrow \ \ \ p=d\frac{3b^2+3b+1}{b^2+b+1} \qquad (2)$ But observe that $\gcd(b^2+b+1, 3b^2+3b+1)=\gcd(b^2+b+1, -3(b^2+b+1)+3b^2+3b+1)=\gcd(b^2+b+1, -2)=1$ Hence $b^2+b+1|d$ and there is a positive integer $c$ such that $d=c(b^2+b+1)$ and equation $(2)$ gives $p=c(3b^2+3b+1)$ Its immediate that $c=1$ and $p=3b^2+3b+1$ . We know that $p=3b^2+3b+1<900$ , so $b \le 16$ . Checking $1\le b\le16$ reveals that the values $b=1, 2, 3, 4, 6, 9, 10, 11, 13, 14$ make $3b^2+3b+1$ a prime, which are $7, 19, 37, 61, 127, 271, 331, 397, 547, 631$ . Thus, answer is $2428$ .

$\\ \boxed{{\color{#D61F06}\text{Second solution. Generalization:}}}$

Let us solve a more general case when the fraction becomes a positive integer like $k$ : $\frac{m^3-n^3}{m^2+n^2-mn‎}=k,$ By the same arguments and notations from first solution we get ‎ $d(a-b)‎\big(a^2+b^2+ab\big)‎=‎k\big( (a-b)^2+ab\big). \qquad (1)$ And there is a positive integer $c$ such that $k=c(a-b)$ , thus ‎ $d\big(a^2+b^2+ab\big) ‎=‎c\big( a^2+b^2-ab\big) . \qquad (2)‎‏‎$ It is clear that $c>d$ . Now we can rearrange this equation as a quadratic in $a:$ $(c-d)a^2-b(c+d)a+b^2(c-d)=0. \qquad (3)‎‏‎$ The discriminant of this quadratic must be a square and that is $\Delta =b^2(c+d)^2-4b^2(c-d)^2=b^2\big((c+d)^2-4(c-d)^2\big).$ Therefore, there is a non-negative integer $e$ such that $‎‎‎(c+d)^2-4(c-d)^2=e^2$ . Setting $e=0$ results in $a=b$ , which is a contradiction. Hence $e$ is positive and the following equation must be solved: $(c+d)^2=4(c-d)^2+e^2.$ But this is Pythagorean equation and its solutions for this case are as follows: $‎‏‎\text{‎(a)‎} \begin{cases} ‎c+d&=f\big(p^2+q^2\big) \\ 2(c-d)&=2fpq \\ e&=f\big(p^2-q^2\big) \end{cases}\qquad ‎‎‏\text{‎(‎‏b)} \begin{cases} ‎c+d&=f\big(p^2+q^2\big) \\ 2(c-d)&=f\big(p^2-q^2\big) \\ e&=‎2fpq‎, \end{cases}$ where positive integers $p>q$ have different parity and are relatively prime, and $f$ is an arbitrary positive integer. ‎In case $(a),$ ‎ $‎\begin{cases} ‎c‎&=\frac{1}{2}f\big(p^2+q^2+pq\big) \\ d&=\frac{1}{2}f\big(p^2+q^2-pq\big) \\ e&=f\big(p^2-q^2\big), \end{cases}‏‎$ and since $p^2+q^2+pq$ and $p^2+q^2-pq$ are both odd, we must have $f=2g$ for a positive integer $g.$ Thus $‎\begin{cases} ‎c‎&=‎g(p^2+q^2+pq) \\ d&=g(p^2+q^2-pq) \\ e&=2g(p^2-q^2), \end{cases}$ ‎ and from equatin $(3)$ $a=\frac{b(c+d)+be}{2(c-d)}=b\frac{p}{q}.$ Note that a negative sign of quadratic formula is not acceptable, because it leads to $a<b$ . Hence $aq=bp,$ and since $\gcd(a, b)=\gcd(p, q)=1$ , we get $a=p$ and $b=q$ and finally ‎‎‎ $(m, n)=(ad, ‎bd)=‎\big(gp(p^2+q^2-pq), gq(p^2+q^2-pq)\big).$ In case $(b),$ ‎ $‎\begin{cases} ‎c‎&=\frac{1}{4}f(3p^2+q^2) \\ d&=\frac{1}{4}f(p^2+3q^2) \\ e&=2fpq. \end{cases}‎‏‎$ Since $3p^2+q^2$ and $p^2+3q^2$ are both odd, we must have $f=4g$ for a positive integer $g$ , and hence $\begin{cases} ‎c&‎=‎g(3p^2+q^2) \\ d&=g(p^2+3q^2) \\ e&=8gpq. \end{cases}$ We get from equation $(3)$ ‎ $a=\frac{b(d+c)+be}{2(c-d)}=b\frac{p+q}{p-q}.$ Hence $a(p-q)=b(p+q)$ and since $\gcd(a, b)=\gcd(p+q, p-q)=1$ , we get $a=p+q$ and $b=p-q$ and finally ‎‎‎ $(m, n)=(ad, ‎bd)=‎\big(g(p+q)(p^2+3q^2), g(p-q)(p^2+3q^2)\big).$ Therefore, we have two family of solutions in general ‎ $‎\begin{cases} ‎(m‎, n)=\big(gp(p^2+q^2-pq), gq(p^2+q^2-pq)\big) \\ ‎(m ‎,n)=‎\big(g(p+q)(p^2+3q^2), g(p-q)(p^2+3q^2)\big), \end{cases}$ where positive integers $p>q$ have different parity and are relatively prime, and $g$ is an arbitrary positive integer. Now, after substitution in the original fraction, these families of solutions lead to two forms for $k$ : ‎ $‎\begin{cases} ‎k=g(p^3-q^3) \\ ‎k=2gq(3p^2+q^2), \end{cases}$ Observe that $‎k=2gq(3p^2+q^2)$ is always a composite number. Therefore, the only possible case is $k=g(p^3-q^3)$ . Notice that in order for $k=g(p^3-q^3)=g(p-q)(p^2+q^2+pq)$ to be a prime number, we must have $g=p-q=1$ . It follows that $k=3q^2+3q+1$ must be a prime number. This is what we had in first solution!

Re-write the following equation as \ $[\frac{m^3 - n^3} {m^3 + n^3}$ ](m+n)= $p$ Now for $p$ to be prime the numerator in the fraction must be a prime. So $m^3 - n^3$ is prime. This implies, $m-n=1$ . Also assume that $gcd(m,n)=d$ . So, $m=ad, n=bd$ with a>b; and $gcd( a,b)=1$ . Now proceed as the same way as before...