I recently came onto a set of numbers with an interesting pattern, but I can't prove it always holds true. I'd appreciate any help in figuring it out.

Let the set $S$ $$S = \left \{ \frac{1}{q^2+1} , q \in \mathbb{Q^+} \right \}$$

And three elements of that set $a,b,c \in S$

My conjecture is that if their average is equal to $\frac{1}{2}$, then at least one of them must be equal to $\frac{1}{2}$.

$$\frac{a+b+c}{3} = \frac{1}{2} \implies (a =\frac{1}{2}) \lor (b=\frac{1}{2}) \lor (c=\frac{1}{2})$$

Do you have any idea on how to prove, or disprove that claim?

Edit to provide examples: You can easily generate sets of numbers that verify the conjectured pattern. Let $f(q) = \frac{1}{q^2+1}$ for rational $q$.

All triplets of the form $(f(1), f(q), f(\frac{1}{q}))$ have an average of $\frac{1}{2}$.

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    $\begingroup$ Could you comment on what numerical testing you've seen that satisfies the conjecture? Like the number of triplets or magnitudes of numbers? $\endgroup$ Jul 31 at 10:20
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    $\begingroup$ Will you please explain what this has to do with elliptic-curves? $\endgroup$ Jul 31 at 10:23
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    $\begingroup$ Are there nontrivial examples of such triples? Trivial being $q_1=q_2=q_3 = 1$. $\endgroup$
    – AlvinL
    Jul 31 at 10:24
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    $\begingroup$ @JuanMoreno He means $(f(1), f(q), f(1/q) ).$ $\endgroup$ Jul 31 at 11:03
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    $\begingroup$ @MafPrivate That's where I started from but I had no luck going in that direction, sadly. It feels like there should be something interesting to investigate though! $\endgroup$
    – Pyrofoux
    Jul 31 at 11:46

2 Answers 2


Unfortunately, the postulated conjecture is not true:

Indeed, examples of admissible $q$-triples that disprove the statement are

$$ (\tfrac{3}{2}, \tfrac{2}{9},\tfrac{41}{23})\\ (\tfrac{9}{2}, \tfrac{2}{3},\tfrac{23}{41}) $$

I found those numbers by just brute force looking for them. I did this by iterating through rational numbers $a,b$, then I calculate $c$ such that $f(a)+f(b)+f(c)=\tfrac{3}{2}$ and then check if $c$ is rational. Please have a look at the following Maple code




for p1 from 2 to N do
  for q1 from 1 to p1-1 do
    for p2 from 1 to N do
      for q2 from p2+1 to N do

        if( type(c, rational) and R<>1/2 ) then 
        end if;

      end do;
    end do;
  end do;
end do;

Increasing $N$ in the code produces more and more triplets that violate the conjecture.

  • $\begingroup$ Excellent! Can you find a reason why the second triple contains the inverses of the elements of the first triple? $\endgroup$ Aug 2 at 19:14
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    $\begingroup$ @Gribouillis $\tfrac{1}{(q^{-1})^2+1} = \tfrac{q^2}{q^2+1} = 1 - \tfrac{1}{q^2+1}$. So if $\sum_{i=1}^m \tfrac{1}{q_i^2+1} = \tfrac{m}{2}$, then $\sum_{i=1}^m\tfrac{1}{(q^{-1})^2+1} = m-m/2 = m/2$ as well. $\endgroup$ Aug 2 at 19:22
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    $\begingroup$ @maxmilgram Oh, excellent! Thank you for sharing your code. I'll resolve the answer and give you the bounty. I'm wondering though, because it's heavily tied to the problem I was initially researching: did you find any solution where the the 3 numbers are $\leq \sqrt{2}$ ? $\endgroup$
    – Pyrofoux
    Aug 2 at 21:04
  • $\begingroup$ Do you mean the numbers $\tfrac{1}{1+x^2}$? $\endgroup$
    – maxmilgram
    Aug 2 at 21:48
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    $\begingroup$ $(\tfrac{79}{61},\tfrac{23}{43},\tfrac{2777}{2033})$ seems to be such a triple. $\endgroup$
    – maxmilgram
    Aug 2 at 21:57






The ratios $x,y,z$ satisfy this relationship: $(1-z^2)(1+x^2)(1+y^2)=2(1+z^2)(x^2y^2-1)$.

Further, this relationship holds when pair wise swapping any of the ratios.

So as pointed out, if any 1 of them is 1, the other two must be reciprocals.



So long as $xy\ne1$, $z\ne 1\to \frac{1}{1+z^2}\ne 1/2$. One still needs to demosntrate existence.

Let $x=c/d, y=e/f$ and $gcd(c,d)=gcd(e,f)=1$






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