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?
 A: 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.
A: 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.
