I Like 315 Irrationally!

Rewrite the equation as $(m-n)(m+n)^2 = 315^9$ Notice that $m-n=a$ must be a positive integer. We get that $(m+n)^2 = \frac{315^9}{a}=\frac{35 \times (3 \times 315^4)^2}{a}.$ Thus $a= 35b^2$ for $b \in \mathbb{N}$ . We deduce that $m-n \; = \; 35b^2 \qquad \qquad m+n \; = \; \frac{3 \times 315^4}{b}$ Solve for $m$ and $n$ : $m \; = \; \frac12\left(\frac{3 \times 315^4}{b} + 35b^2\right) \qquad \qquad n \; = \; \frac12\left(\frac{3 \times 315^4}{b} - 35b^2\right)$ $n$ must be positive, it follows that $(3 \times 315^4)/b > 35b^2$ , and $35b^3 < 3 \times 315^4$ . Finally $b^3 < 3^3 \times 315^3 \ \ \ \ \Longrightarrow \ \ \ \ b<3\times 315=945 \ \ \ \ \Longrightarrow \ \ \ \ 1 \le b \le 944$ Each positive integer $b \le 944$ corresponds to a solution pair for $(m, n)$ . Thus the answer is $\boxed{944}$ .

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Comment:

Why $(m+n)^2 = \frac{315^9}{a}=\frac{35 \times (3 \times 315^4)^2}{a}$ implies that $a= 35b^2$ for $b \in \mathbb{N}$ ?

There are many ways to explain this. I prefer this one: We have $\frac{35}{a}= \left( \frac{m+n}{3 \times 315^4} \right)^2=$ square of a rational number. Thus we can write $\frac{35}{a}=\frac{c^2}{b^2}$ for positive integers $b$ and $c$ , where $\gcd(b, c)=1$ .

Now, we get $35b^2=ac^2$ . It follows that $c^2 \mid 35 b^2$ and $c^2 \mid 35$ since $b$ and $c$ are co-prime. But $1$ is the only perfect square dividing $35$ . Thus, $c=1$ and $a=35b^2$ .

$m^3-n^3+mn(m-n) = 315^9$

$(m-n)(m^2+mn+n^2)+mn(m-n) = 315^9$

$(m-n)(m^2+2mn+n^2) = 315^9$

$(m-n)(m+n)^2 = 315^9$

Since $\frac{315^9}{(m+n)^2} > 0$ , we have $m-n>0$ .

Let $s=m-n$ where s is a positive integer. Plugging it into the equation:

$s(2n+s)^2 = 315^9 = (3^2)^9 \cdot 35^9$

$(\frac{2n+s}{3^9})^2 = \frac {35^9}{s}$

It follows that $\frac{35^9}{s}$ is a square of a rational, thus $s$ must be in the form $35k^2$ for a positive integer $k$ . Hence our equation becomes

$(\frac{2n+35k^2}{3^9})^2 = \frac{35^8}{k^2}$

$\frac{2n+35k^2}{3^9} = \frac{35^4}{k}$

$n = \frac{3^9 \cdot 35^4 - 35k^3}{2k}$

$m = n+s = \frac{3^9 \cdot 35^4 + 35k^3}{2k}$

From $n>0$ we obtain

$\frac{3^9 \cdot 35^4 - 35k^3}{2k} > 0$

$3^9 \cdot 35^3 > k^3$ $\Rightarrow$

$1 \leq k < 3^3 \cdot 35 = 945$ .

For such values of $k$ both $m$ and $n$ will be positive rationals hence solutions of our equation. Since $k$ can have 944 possible values and for different $k$ 's there are different pair of solutions, our answer is 944.

Rewrite the equation as $(m-n)(m+n)^2=315^9$ We are told $m-n\in\mathbb{Z}$ , but clearly $m-n>0$ so $m-n\in\mathbb{N}$ . There is then clearly a bijection between solutions to the original problem and the solutions to $a(a+2n)^2=315^9$ for $a\in\mathbb{N}$ and $n\in\mathbb{Q}$ , by letting $a=m-n$ . Since $n$ is determined by our choice of $a$ , the pair $(a,n)$ is a solution iff $a\in\mathbb{N}$ and $n=\frac{1}{2}\left(\sqrt{\frac{315^9}{a}}-a\right)\in\mathbb{Q}\quad\text{and}\quad n=\frac{1}{2}\left(\sqrt{\frac{315^9}{a}}-a\right)>0$ The second part coming from the constraint that $n$ be positive. The second part is equivalent to $a< 31255875$ . We also have $\frac{1}{2}\left(\sqrt{\frac{315^9}{a}}-a\right)\in\mathbb{Q}\Leftrightarrow \sqrt{\frac{315^9}{a}}\in\mathbb{Q}\Leftrightarrow \sqrt{\frac{5\cdot 7}{a}}\in\mathbb{Q}\Leftrightarrow a=5\cdot 7\cdot b^2$ for some $b\in\mathbb{N}$ . Thus $(a,n)$ is a solution iff $a< 31255875$ and $a=5\cdot 7\cdot b^2$ for some $b$ , which gives us the freedom to choose any $1\le b< 945=\sqrt{31255875/(5\cdot 7)}$ .