# Can an infinite sequence of integers generate integer-area triangles?

(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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uhm, 345345345... ? We probably need unbounded or something. – Hailong Dao Jan 17 '10 at 8:38
The question has been edited and fixed to reflect the original intent of the problem. – Harry Gindi Jan 18 '10 at 0:33
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). – Qiaochu Yuan Jan 18 '10 at 1:13
Agree with Qiaochu: no obvious reason why it has to be increasing. – Hailong Dao Jan 18 '10 at 5:50
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". – Kevin Buzzard Jan 19 '10 at 22:24

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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apart from (0,1), I guess – Mariano Suárez-Alvarez Jan 20 '10 at 1:23
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. – Bjorn Poonen Jan 20 '10 at 2:21
Thanks Bjorn! Looks like I need a better idea. – David Speyer Jan 20 '10 at 3:09
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. – Kevin Buzzard Jan 20 '10 at 22:01
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. – Bjorn Poonen Jan 22 '10 at 2:54

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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It's not clear to me why such a sequence would have to have that sort of periodicity. – Alison Miller Jan 19 '10 at 19:27
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. – David Speyer Jan 19 '10 at 21:21
updating to reflect the possibility of lack of periodicity – Sparr Jan 19 '10 at 22:11
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. – Sparr Mar 2 '10 at 6:29