TITLE: Can an infinite sequence of integers generate integer-area triangles?
QUESTION [15 upvotes]: (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?

REPLY [4 votes]: 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.