Rational Solutions of the Diophantine Equation: y² = x³ - 2?

Favorite Answer

I just ran a computer program to find solutions other than x=3,y=5 and x=1290/1000, y=383/1000, for denominators ≤ 60³ and did not find any. I looked for integer solutions to y² - x³ + 2n^6 = 0, where n³ would be the denominator, as in

(y/n³)² = (xn/n³)³ - 2

Addendum: It was a simple, brute force program, with 3 nested "For" loops, beginning a For loop for n from 1 to 100, followed by another For for y from 1 to 1000 and For for x from 1 to 1000. It took the better part of an hour, and this was just a preliminary trial to get a feel for it. I can run a more exhaustive trial overnight. Oh yeah, it was included in the coding to disregard the 2 known solutions.

By the way, number theory is one of my weak subjects. I am not well versed in Diophantine theory.

Addendum 2: Using a slightly modifed computer search method, there exists no other solutions for any x' < 10,000, where x = x'n, and n is the denominator. I can push it up to x' < 100,000, but it would probably take all night. I doubt if I'll find anything, given the extreme scarcity of solutions.

Addendum 3: After fidding around with this problem, I have found another rational solution which is as follows:

(66234835 / 5000211)² = (164323 / 29241)³ - 2

Addendum 4: I made use of an extremely interesting property of elliptic curves which is that if any line intersects it in 3 places, and 2 of them are rational, the 3rd is rational also. Remember that this elliptic curve has both positive and negative values (because of y²), so I picked the points (129/100, -383/1000), (3,5) instead of (129/100, 383/1000), (3,5), because the latter's first point is actually a tangent, so that there is no 3rd point. It's straightfoward to find the equation for the straight line, and then make it equal to √(x³ -2), ending up with a cubic equation with massive rational coefficients. But the wonderful thing is, you already know two of the factors---the other two solutions! (x - 3) and (x - 129/100). Voila, you're left with (x - 164323 / 29241), and the rest follows. So, I learned something new today about elliptic curves, even as I knew they're supposed to have a lot of interesting properties.

Addendum 5: The property described above would be true for any polynomial curve that would preserve the degree of the polynomial to be solved to find such points of intersection. And in this case, I can keep drawing more such lines and get Delauney-like solutions, as for yet another example:

(15230044576037327107200537568897 / 11623520729117946174953656293000)² = (794845361623184880769 / 513127310073606144900)³ - 2...Show more

Interesting. Before i look at the links some observations:
Set y=p/q lowest terms, x=r/s lowest terms, then
p^2s^3 = r^3q^2 - 2q^2s^3 gives s^3 | q^2, q^2 | s^3
and s^3 = q^2 = t^6, giving a diophantine equation

(*)............. p^2 = r^3 - 2t^6. For which you have found

p=383, r=129, t= 10. The values for the denominators
are generally q=t^3 and s=t^2 ; t=1 gives p=5,r=3.
At this point i don't know about 1<t<10. Visit later
ty for the problem. Since i have no computer programs, i
will have to see what i can get formally from (*):

r^3 = p^2+2t^6 = (p+t^3 * 2^.5*i)(p-t^3*2^.5*i)
= (a + b*2^.5*i)^3*(a - b*2^.5*i)^3.

Equating real and imaginary coefficients we get,

p = a^3 - 6ab^2

r = a^2 + 2b^2

t^3 = 3a^2b - 2b^3 = b(3a^2 - 2b^2)

for some integers a and b (necessarily coprime since
(p,r) =1 ) forcing b =c^3 and (3a^2 - 2b^2) = d^3
again for integers c and d. Recall that t^3 is also
the larger of the denominators in the original equation
with the rational numbers p/q and r/s, the smaller being
t^2.
This analysis may or may not be completable by these
elementary methods but the formulas for p,r,t^3 may
help to characterize the computer search. If more formal
stuff occurs to me i'll post it later. Just a note:
We can recast your solution:

x=1290/1000 and y =383/1000 in the form,

{(3(128) - 1)/ 10^3 }^2 = {(128 + 1)/ 10^2 }^3 - 2

for t = 10....Show more

Joey is essentially correct, but he made a calculation error. The correct parameterization is x = u^2 + 2uv -v^2 y = -u^2 + 2uv +v^2 z = u^2 + v^2. We also need x,y,z > 0 somehow. Clearly, z>0. Then we need u^2 + 2uv - v^2 > 0 <--> (u+v)^2 > 2v^2 <--> |u+v| > sqrt(2)*|v|, and -u^2 + 2uv +v^2 > 0 <--> (u+v)^2 > 2u^2 <--> |u+v| > sqrt(2)*|u|. You should also be able to conclude that u and v are either both positive or both negative (add together the far left inequalities from the two lines above). Furthermore, either A) one of u and v is even and the other odd with gcd(u,v) = 1; or B) u = u'/sqrt(2) and v = v'/sqrt(2), where u' and v' are both odd with gcd(u',v') = 1. To see part B, try u = 3/sqrt(2) and v = 5/sqrt(2), they yield a positive integer solution (x,y,z). I don't show the work for the last part, but you should be able to show that using two odds or two evens yield all even (x,y,z), and the (u,v) given in part B yield an odd integer (x,y,z) solution. Good luck!...Show more