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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 1000 and didn't find any, but I don't think that is large enough to even count as evidence against one existing.

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.

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 numbers 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.

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 1000 and didn't find any, but I don't think that is large enough to even count as evidence against one existing.

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

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

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 numbers 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.

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 numbers 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.