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(asked by Shanzhen Gao, shanzhengao at yahoo.com, on the Q&A board at JMM)

Does there exist an infinite, monotonically increasing sequence of integers $\{ a_n \}_{n \geq 0}$ such that for any $n$, the three integers $(a_n, a_{n+1}, a_{n+2})$ are the side lengths of a plane triangle with integer area?

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    $\begingroup$ uhm, 345345345... ? We probably need unbounded or something. $\endgroup$
    – Hailong Dao
    Commented Jan 17, 2010 at 8:38
  • $\begingroup$ The question has been edited and fixed to reflect the original intent of the problem. $\endgroup$
    – Harry Gindi
    Commented Jan 18, 2010 at 0:33
  • $\begingroup$ This is unlikely, but it could happen that such a sequence exists which is unbounded but essentially non-monotonic (i.e. it remains non-monotonic even after finitely many terms are omitted). $\endgroup$
    – Qiaochu Yuan
    Commented Jan 18, 2010 at 1:13
  • $\begingroup$ Agree with Qiaochu: no obvious reason why it has to be increasing. $\endgroup$
    – Hailong Dao
    Commented Jan 18, 2010 at 5:50
  • $\begingroup$ I can't answer this question. I think I'd have more chance if I were allowed to use rational numbers rather than integers. My computer says "5863, 14820, 19825, 29575, 32500, 51675, 54575". $\endgroup$
    – Kevin Buzzard
    Commented Jan 19, 2010 at 22:24

2 Answers 2

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I'll throw out a dumb idea: can anyone find a rational point on

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

UPDATE: The above formula used to have a sign error, which I have just fixed, and Bjorn's reponse was to the version with the sign error. Thanks to Kevin Buzzard for pointing this out to me.

Because, if so, $a_n=x^n$ gives triangles with rational area. Of course, this still wouldn't give an integer solution, but it would rule out a number of easy arguments against one existing.

I did a brute force search of values of $x$ with numerator and denominator under 5000 and didn't find any, but I don't think that is large enough to even count as evidence against one existing.

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  • $\begingroup$ apart from (0,1), I guess $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 20, 2010 at 1:23
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    $\begingroup$ There are no rational points other than $(0,\pm 1)$. Your curve has the form $y^2=f(x^2)$. So it maps to the genus 1 curve $y^2=f(x)$, whose smooth projective model has 4 obvious rational points (above $x=0$ and $x=\infty$). There are no others since Magma says that its Jacobian (which is the same curve) has rank 0 and torsion subgroup of size 4. $\endgroup$
    – Bjorn Poonen
    Commented Jan 20, 2010 at 2:21
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    $\begingroup$ Hold your horses everyone. Hasn't David made a sign error? y^2=-RHS is the relevant equation, right? @David: I don't think this will pan out, even with the sign fix. However trying something with small period bigger than 1 is probably worth doing. $\endgroup$
    – Kevin Buzzard
    Commented Jan 20, 2010 at 22:01
  • $\begingroup$ Update: David fixed the sign error. Thanks David. So now Bjorn's comments don't apply because unfortunately the genus 1 curve y^2=f(x) has (for me at least) no obvious rational points at all. There are standard ways of analysing this curve but we have to wait until someone applies them. First thing to do is perhaps to check for no p-adic points for primes p of bad reduction. $\endgroup$
    – Kevin Buzzard
    Commented Jan 21, 2010 at 17:18
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    $\begingroup$ The corrected curve has no $2$-adic points. In fact, even the genus 1 curve it maps to has no $2$-adic points. Proof: The quartic is $-1$ mod 4 whenever $x$ is in $\mathbb{Z}_2$. Because of the $x \mapsto 1/x$ symmetry, there aren't any points with $x \in \mathbb{Q}_2$ either. $\endgroup$
    – Bjorn Poonen
    Commented Jan 22, 2010 at 2:54
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To find this sequence, if it exists, you would need to find a set of Heronian triangles such that the lengths of sides a:b:c in each triangle corresponded to sides b:c:d in the next. The series of triangle proportions could (and likely will, if it exsts, I think) contain a cycle wherein a multiplier is introduced after each cycle, such that the triangles after x:y:z are y:z:(a*n), z:(a*n):(b*n), and (a*n):(b*n):(c*n).

A cursory search of the hundred smallest integer Heronian triangles yields no such set longer than 2 triangles, and no such cycle. As Heronian triangles can be parametrically enumerated, it would be possible to perform a brute force search of a sizable number of them for such a sequence or cycle.

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  • $\begingroup$ It's not clear to me why such a sequence would have to have that sort of periodicity. $\endgroup$
    – Alison Miller
    Commented Jan 19, 2010 at 19:27
  • $\begingroup$ I don't see why either. On the other hand, this is a plausible way that one could construct such a sequence, so it is worth recording that it doesn't work. $\endgroup$
    – David E Speyer
    Commented Jan 19, 2010 at 21:21
  • $\begingroup$ updating to reflect the possibility of lack of periodicity $\endgroup$
    – Sparr
    Commented Jan 19, 2010 at 22:11
  • $\begingroup$ Also, to clarify, while periodicity is not a requirement, I think a periodic solution is far more likely, as it would require a smaller number of non-congruent triangles to be found. $\endgroup$
    – Sparr
    Commented Mar 2, 2010 at 6:29

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