# If 3 numbers of the form $\frac{1}{q^2+1}$ have a mean of $\frac{1}{2}$, will one of them always be $\frac{1}{2}$?

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 $$\frac{1}{2}$$.

• Could you comment on what numerical testing you've seen that satisfies the conjecture? Like the number of triplets or magnitudes of numbers? Jul 31 at 10:20
• Will you please explain what this has to do with elliptic-curves? Jul 31 at 10:23
• Are there nontrivial examples of such triples? Trivial being $q_1=q_2=q_3 = 1$. Jul 31 at 10:24
• @JuanMoreno He means $(f(1), f(q), f(1/q) ).$ Jul 31 at 11:03
• @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! Jul 31 at 11:46

## 2 Answers

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

restart:

f:=x->1/(1+x^2);

N:=10;

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

a:=p1/q1;
b:=p2/q2;
R:=3/2-f(a)-f(b);
c:=sqrt(1/R-1);

if( type(c, rational) and R<>1/2 ) then
print(a,b,c,f(a),f(b),f(c),f(a)+f(b)+f(c))
end if;

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


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

• Excellent! Can you find a reason why the second triple contains the inverses of the elements of the first triple? Aug 2 at 19:14
• @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. Aug 2 at 19:22
• @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 $\leq \sqrt{2}$ ? Aug 2 at 21:04
• Do you mean the numbers $\tfrac{1}{1+x^2}$? Aug 2 at 21:48
• $(\tfrac{79}{61},\tfrac{23}{43},\tfrac{2777}{2033})$ seems to be such a triple. Aug 2 at 21:57

$$\frac{1}{1+x^2}+\frac{1}{1+y^2}=\frac{x^2+y^2+2}{(1+x^2)(1+y^2)}$$

$$\frac{3}{2}-\frac{x^2+y^2+2}{(1+x^2)(1+y^2)}=\frac{3(1+x^2)(1+y^2)-2(x^2+y^2+2)}{2(1+x^2)(1+y^2)}$$

$$=\frac{3+3x^2+3y^2+3x^2y^2-2x^2-2y^2-4}{2(1+x^2)(1+y^2)}=\frac{x^2+y^2+3x^2y^2-1}{2(1+x^2)(1+y^2)}$$

$$\frac{1}{1+z^2}=\frac{x^2+y^2+3x^2y^2-1}{2(1+x^2)(1+y^2)}$$

$$2(1+x^2)(1+y^2)=(1+z^2)[(1+x^2)(1+y^2)+2x^2y^2-2]$$

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.

$$(1+x^2)(1+y^2)-2(x^2y^2-1)=2z^2(x^2y^2-1)+z^2(1+x^2)(1+y^2)$$

$$z^2=\frac{(1+x^2)(1+y^2)-2(x^2y^2-1)}{2(x^2y^2-1)+(1+x^2)(1+y^2)}$$

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 $$gcd(c,d)=gcd(e,f)=1$$

$$\frac{p^2}{q^2}=\frac{(c^2+d^2)(e^2+f^2)-2(c^2e^2-d^2f^2)}{2(c^2e^2-d^2f^2)+(c^2+d^2)(e^2+f^2)}$$

$$2p^2(c^2e^2-d^2f^2)+p^2(c^2+d^2)(e^2+f^2)=q^2(c^2+d^2)(e^2+f^2)-2q^2(c^2e^2-d^2f^2)$$

$$2(p^2+q^2)(c^2e^2-d^2f^2)+(p^2-q^2)(c^2+d^2)(e^2+f^2)=0$$

$$2(p^2+q^2)(d^2f^2-c^2e^2)=(p^2-q^2)(c^2+d^2)(e^2+f^2)$$