An equation of fractional parts

It is clear that any $1 \le n \le 15$ will do, since we can choose $m=15$ .

Suppose then that $m \ge n \ge 16$ . I first claim that the result holds for any $m \ge n \ge 16$ with $\tfrac{1}{15} = \tfrac{1}{m} + \tfrac{1}{n}$ . If this identity holds, then $\left\lfloor \tfrac{x}{15}\right\rfloor - \left\lfloor \tfrac{x}{n} \right\rfloor - \left\lfloor \tfrac{x}{m}\right\rfloor + \left\{\tfrac{x}{15}\right\} - \left\{\tfrac{x}{n}\right\}\; = \; \left\{ \tfrac{x}{m}\right\}$ If $\left\{\tfrac{x}{15}\right\} - \left\{\tfrac{x}{n}\right\} \ge 0$ , it is clear that it is equal to $\left\{\tfrac{x}{m}\right\}$ . If $\left\{\tfrac{x}{15}\right\} - \left\{\tfrac{x}{n}\right\} < 0$ , it is clear that it is equal to $\left\{\tfrac{x}{m}\right\}-1$ . Thus we deduce that $\left\{\tfrac{x}{15}\right\} - \left\{\tfrac{x}{n}\right\} - \left\{\tfrac{x}{m}\right\} \; = \; 0\,,\,-1 \; \le \; 0$ for all real $x$ .

Solving the equation $n^{-1} + m^{-1} = 15^{-1}$ for $m\ge n \ge 16$ leads to the equation $(m-15)(n-15)=225$ , and so we can have $n=16,18,20,24,30$ , with respective values of $m$ being $240,90,60,40,30$ . I claim there are no other values of $n$ that will do.

Suppose now that $m \ge n \ge 16$ and that $m,n$ satisfy the stated condition. Then (putting $x=1$ ) we deduce that $15^{-1} \le n^{-1} + m^{-1}$ . Hence $225 \; \ge \; (m-15)(n-15) \; \ge \; (n-15)^2$ so that $16 \le n \le 30$ . We also note, if we put $x=\{m,n\}$ to be the lowest common multiple of $m$ and $n$ , that $15$ must divide $\{m,n\}$ .

If $m$ and $n$ are coprime, find integers $a,b$ such that $na-mb=1$ . Putting $x=na$ we deduce that $\left\{\tfrac{an}{15}\right\} \le \tfrac{1}{m}$ . Thus $15$ must divide $an$ . Hence $15$ divides $an^2 - bmn = n$ .

Consider the HCF $(m,n)$ of $m$ and $n$ .

• If $(15,n)=15$ , we must have $n=30$ .

• If $(15,n)=5$ , we must have $n=20,25$ . If $n=25$ , then $m$ would have to be a multiple of $3$ between $25$ and $37.5$ . But $15$ does not divide $n$ , so $n$ and $m$ cannot be coprime. Thus $m$ must be a multiple of $5$ as well, and hence must be $30$ . But putting $x=125$ shows that this cannot work. Thus we deduce that $n=20$ is the only option.

• If $(15,n)=3$ , then $n = 18,21,24,27$ . If $n=27$ then $m$ would be a multiple of $5$ between $16$ and $33.75$ . Since $(n,m) \neq 1$ , $m$ would be a multiple of $3$ as well, and so would have to be $30$ . Putting $x=-27$ shows that does not work. If $n=21$ then $m$ would be a multiple of $5$ between $16$ and $52.5$ . Since $(n,m) \neq 1$ , $m$ would have to be a multiple of either $3$ or $7$ as well, and hence must be one of $30,35,45$ . But $m=30$ does not work (put $x=-21$ ), and $m=35$ does not work ( $x=42$ ), and $m=45$ does not work ( $x=-84$ ). Thus only $n=18,24$ are possible.

• If $(15,n)=1$ , then $m$ would have to be multiple of $15$ between $16$ and $240$ . Thus $m=15k$ where $2 \le k\le 16$ , and $2 \le (n,m) = (n,k) \le 16$ . Note that we cannot have $k=15$ . Find integers $a,b$ with $an + bm = (n,m)$ . If $(n,m)<16$ , putting $x=an$ gives $\tfrac{(n,m)}{15}=\left\{\tfrac{an}{15}\right\} \le \left\{\tfrac{an}{m}\right\} = \tfrac{(n,m)}{m}$ , which is impossible. Thus $(n,m)=16$ , and hence $n=16$ is the only option.

Thus the only possible values of $n$ are $1,2,\ldots,15,16,18,20,24,30$ , and the sum of these numbers is $228$ .

Because the given inequality is satisfied for all real x, we consider two cases:

Case 1: m=15, then the given condition is obviously satisfied, we get 15 possible values of n, from 1 to 15.

Case 2: $m \geq n>15$ .

We have: { $\frac{x}{15}$ } -{ $\frac{x}{m}$ }-{ $\frac{x}{n}$ } $\leq$ { $\frac{x}{15}$ } -{ $\frac{x}{m}+\frac{x}{n}$ }

$\leq$ { $\frac{x}{15} -\frac{x}{m}-\frac{x}{n}$ }.

The given condition is satisfied for all real x when $\frac{1}{15}=\frac{1}{m}+\frac{1}{n}$ .

We get (m,n)=(16,240);(18,90);(20,60);(24,40);(30,30).

The answer is: $\frac{15*16}{2} +16 +18+20+24+30=228$