orl wrote: For a finite set $ X$ of positive integers, let $ \Sigma(X) = \sum_{x \in X} \arctan \frac {1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac {\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) = \frac {\pi}{2}.$ Quite nice problem. $ \tan(\Sigma(S))$ is a rational number and we are looking for a complementary set $ U$ such that $ \tan(\Sigma(U))$ $ =\frac{1}{\tan(\Sigma(S))}$. Then $ T=S\cup U$ would be an answer. So it would be enough to show that, given any positive rational number $ \frac ab$ (with $ \gcd(a,b)=1$), we can always find a set $ U$ such that $ \tan(\Sigma(U))=\frac ab$ If $ a=1$, then $ U=\{b\}$ is the answer. If $ a>1$, let then $ n=1+[\frac ba]$ and $ x=\frac{an-b}{bn+a}$. If we find a set $ V$ such that $ \tan(\Sigma(V))=x$, then $ \tan(\Sigma(V\cup\{n\}))=\frac ab$ and we got the answer. It remains just to see that $ n=1+[\frac ba]$ and $ a>1$ $ \implies$ $ n-1<\frac ba < n$ $ \implies$ $ 0 < an-b < a$ And so the numerator of $ x$ is less than $ a$ and we are sure we can repeat the process a finite number of times to get the result. The end.

pco wrote: orl wrote: For a finite set $ X$ of positive integers, let $ \Sigma(X) = \sum_{x \in X} \arctan \frac {1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac {\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) = \frac {\pi}{2}.$ Quite nice problem. $ \tan(\Sigma(S))$ is a rational number and we are looking for a complementary set $ U$ such that $ \tan(\Sigma(U))$ $ = \frac {1}{\tan(\Sigma(S))}$. Then $ T = S\cup U$ would be an answer. So it would be enough to show that, given any positive rational number $ \frac ab$ (with $ \gcd(a,b) = 1$), we can always find a set $ U$ such that $ \tan(\Sigma(U)) = \frac ab$ If $ a = 1$, then $ U = \{b\}$ is the answer. If $ a > 1$, let then $ n = 1 + [\frac ba]$ and $ x = \frac {an - b}{bn + a}$. If we find a set $ V$ such that $ \tan(\Sigma(V)) = x$, then $ \tan(\Sigma(V\cup\{n\})) = \frac ab$ and we got the answer. It remains just to see that $ n = 1 + [\frac ba]$ and $ a > 1$ $ \implies$ $ n - 1 < \frac ba < n$ $ \implies$ $ 0 < an - b < a$ And so the numerator of $ x$ is less than $ a$ and we are sure we can repeat the process a finite number of times to get the result. The end. No. Elements in set need to be distinct. You only proved this problem for list.

greentreeroad wrote: No. Elements in set need to be distinct. You only proved this problem for list. You are perfectly right and I thank you for having noticed this. This can be rather easily solved using the fact that $ \arctan(\frac 1p)=\arctan(\frac 1{p+1})+\arctan(\frac 1{p^2+p+1})$ So we can replace in $ U$ any "bad" element $ p$ by the couple $ \{p+1,p^2+p+1\}$ so to transform the list in set (with $ U\cap S=\emptyset$). This process of replacement need to be conducted with some caution in order to avoid creating infinitely many double values, but it's rather easy to find such a process.

pco wrote: So we can replace in $ U$ any "bad" element $ p$ by the couple $ \{p + 1,p^2 + p + 1\}$ so to transform the list in set (with $ U\cap S = \emptyset$). This process of replacement need to be conducted with some caution in order to avoid creating infinitely many double values, but it's rather easy to find such a process. How does this process work precisely? What are the invariants? It is not obvious to me.

We can modify pco's proof in the following way: say the sum we have is arctan(a/b), then you want to come up with some more denominators such that the arctan sum becomes arctan(b/a). We can keep subtracting the largest possible fraction 1/n<b/a that is not in S. Clearly, the arctan(b'/a') value left will always have b'/a'>0 (or 0, in which we're done), and we basically keep going until we've cleared all the elements of S (we're subtracting smaller fractions than all the ones in S). It's not too hard to show that (1) we won't subtract the same fraction twice and (2) we'll eventually hit a point where all the fractions we take are such that 1/n<b/a<1/(n-1). From here, the numerators will decrease. (We'll need the observation that arctan(1/n)~1/n)

For the sake of completeness... test20 wrote: How does this process work precisely? What are the invariants? It's basically the same as the proof here, except you do $n\to\{n+1,n^2+n+1\}$ instead of $n\to\{n+1,n^2\}$. BTW, the same thing (with $n\to\{n+1,n(n+1)\}$) works for Egyptian fraction representations as well (but so does Catalyst's greedy algorithm).

pco your "easy process" is actually very hard and not at all trivial Best thing for this problem is Catalyst's

Copied from my solution in Math StackExchange. (I indeed know that this is pretty much a rewording of pco's proof) I will prove a stronger statement - the RMM problem itself. It generalizes the two integers $x,y$ to a set of positive integers. First, for convenience, we define $f(x)=\arctan x$. Also, for a set $X$, we define $g(X)=\sum_{x \in X} f(\frac{1}{x})$. Now, say we are given a set $X$, where $g(X)=f(\frac{a}{b}) < \frac{\pi}{2}$ and $(a,b)=1$. It suffices to find a set $Y$, such that $X \cap Y = \emptyset$ and $g(Y)=f(\frac{b}{a})$, since we will have $$g(X \cup Y)=g(X)+g(Y)=f(\frac{a}{b})+f(\frac{b}{a})=\frac{\pi}{2}$$ Disregard the fact that we need distinct positive integers, for now. Now if $b=1$, we are done since we can just let $Y=\{a\}$ to finish the problem. Assume that $b \ge 2$. Now set an integer $l \in Y$ which will be determined later. We have $$g(Y-\{l\})=g(Y)-f(\frac{1}{l}) = f(\frac{b}{a})-f(\frac{1}{l})=f(\frac{bl-a}{al+b})$$We want $0<bl-a<b$, which motivates us to set $l=1+\lfloor \frac{a}{b} \rfloor$, since $a$ cannot be a divisor of $b$. Now the numerator strictly decreases, since $\frac{a}{b}<l<1+\frac{a}{b}$. Therefore, after performing this finitely, we will have the numerator equal to $1$, i.e. $b=1$, a case we already had a look on. NOTE: If this is uncomfortable, use strong induction on the numerator. Induction Proof wrote: The base case $b=1$ is the same. Assume that there is a set $P$ such that $g(P)=f(\frac{x}{y})$ if $x<b$. Since $0<bl-a<b$, by induction hypothesis, we have a set $P$ such that $g(P)=f(\frac{bl-a}{al+b})$. Add the number $l$ to find $g(P \cup \{l\})=f(\frac{b}{a})$, which completes the induction. We have determined set $Y$, but we don't know that it has a repeating number. Perform the map $t \mapsto \{t+1,t^2+t+1\}$ on the elements of $Y$. Since the functions $p(x)=x+1$ and $q(x)=x^2+x+1$ are not bounded above, we can perform it finitely to replace the repeating numbers. To make this argument more rigorous, order the elements by size and start from the smallest number. When encountered with a repeating number(s), keep performing the map on the resulting numbers until the numbers that result from this map are larger than any other element. Since the number of the elements of the original set is finite, the number of times that this map needs to be used is finite as well, so the resulting set is finite. We are done. $\blacksquare$