Let S be the set of rational numbers x such that x 3 − 4 3 2 is the square of a rational number.
If S is infinite, enter − 1 . If S is finite, enter the sum of the elements of S .
Here is a very helpful hint:
If
y
2
=
x
3
−
4
3
2
,
let
u
=
6
x
3
6
−
y
,
v
=
6
x
3
6
+
y
. What is
u
3
+
v
3
?
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@Patrick Corn , can we always find a substitution of u and v (as rational functions of y and x ), where y 2 = x 3 − r , for another rational r ?
E.g.: I don't know how to formulate u and v for y 2 = x 3 − 1 1 . And yes, I was thinking about solving this problem by using your hint here.
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No, pretty much only for this specific r (up to multiplication by a perfect sixth power). The family of elliptic curves y 2 = x 3 − D is much, much more interesting than the one Fermat curve u 3 + v 3 = 1 . Many such curves have lots of points--in fact, there are lots of open questions about the structure of those points, including one of the Millennium prize problems .
I highly recommend Joe Silverman's book The Arithmetic of Elliptic Curves . Lots of down-to-earth examples, and a good overview of the theory. (He also has a book for undergraduates called Rational Points on Elliptic Curves , which I don't know as much about but I assume is also good and probably significantly easier.)
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Thanks for your detailed response. I'll try to get the book as well. =D
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Using the very helpful hint (thanks, Patrick!), let
u = 6 x 3 6 − y , v = 6 x 3 6 + y
Then
u 3 + v 3 = x 3 4 3 2 + y 2
so that we immediately have, as a consequence
u 3 + v 3 = 1
which has no known rational solutions per Fermat's Last Theorem , other than
( u , v ) = ( 1 , 0 ) or ( u , v ) = ( 0 , 1 )
which leads to the only rational solutions
( x , y ) = ( 1 2 , − 3 6 ) or ( x , y ) = ( 1 2 , 3 6 )
and so the sum of the elements of S is just 1 2