What are the integer solutions of x^2 -2 = 2 y^3?

It seems to be the largest solution.

x² - 2 = 2 y³

<=>

x² = 2y³ + 2

<=>

4x² = 8y³ + 8

<=> (Setting x=2x' and y=2y' and renaming to x,y for convenience)

x² = y³ + 8

This is, of course, is an elliptic curve. In the link below elliptic curves are listed of this particular form.

The integer points given for our curve are:

(-2,0) , (1,+-3) , (2,+-4) , (46,+-312)

which correspond to solutions for our problem:

(0,-1) , [no integers] , (+-2,1) , (+-156,23)

i) Transferring the given one, x^2 = 2(y^3 + 1)

ii) Since (x,y) are integers, right side of the above is even. ==>x must be even.

Hence, let x be 2n, where n is an integer.

==> y^3 = [(2n)^2 - 2]/2 = 2(n^2) - 1

==> y is an odd integer.

As well from the first one, for x to be real integer, y^3 + 1 must be greater than or equal to zero.

Thus y is an odd integer greater than or equal to -1.

iii) So from the above, x = ±(2n), where n is an integer

y = (2n² - 1)^(1/3) and y ≥ -1

Satisfying the above constraints, some set of integer solutions are:

(x,y) = (0, -1), (±2, 1), (±156, 23)

EDIT:

The last step in point (ii) was, y ≥ 0, which is now corrected as y ≥ -1

Your curve is an elliptic curve. In fact, as Michael points out, it is equivalent to a "Mordell equation" or Mordell elliptic curve, i.e. an elliptic curve of the form Y^2 = X^3 + k., where in your case k=8. You get this by putting y=X/2, x=Y/2 and multiplying by 4. Integer points on Y^2 = X^3 + 8 are either integer or half-integer points on your curve. These Mordel curves have been very well studied. In the old days (e.g. 1970's) papers were written on how to find all integer points, for certain specific k's. See for example:

http://www.sciencedirect.com/science/article/pii/0...

http://www.sciencedirect.com/science/article/pii/0...

(I have no access to a good library, nor to jstor, so can only see the abstracts).

A method called "Baker's effective method" is used in one of those (I don't know about it).

Possibly ad hoc elementary methods might sometimes work, see for example: http://www.math.uconn.edu/~kconrad/ross2008/mordel...

Nowadays, people use software such as magma or the free one sage to answer the question of finding all integer points (or S-integer points, which I think have denominators powers of some prime ??) on elliptic curves. I think that sage can certify (presumably based on theorems) whether all integer points have been found. (If you are willing to trust software which likely has bugs). Apparently Michael found a table for Mordell curves, indicating that (156,23) is the largest integer point on your curve. Maybe sage generated the table.

It would be really nice to have an elementary proof that (156,23) is largest, and one might be possible, but I don't have it.

The above is my impression, but largely I am ignorant about this, caveat lector.

Edit: The table Michael refers to was apparently not generated by sage, but rather by SIMATH. But see:

http://tnt.math.se.tmu.ac.jp/simath/

Edit2: The state of the art for Mordell's equation as of 2009 is given in this paper: http://www.inf.unideb.hu/~pethoe/cikkek/67_MORDELL... Unfortunately the online version is missing the graphs.

x^2 - 2 = 2 * y^3

x^2 = 2 * y^3 + 2

x^2 = 2 * (y^3 + 1)

x^2 = 2 * (y + 1) * (y^2 - y + 1)

2 * (y + 1) = y^2 - y + 1

2y + 2 = y^2 - y + 1

0 = y^2 - 3y - 1

y = (3 +/- sqrt(9 + 4)) / 2

y = (3 +/- sqrt(13)) / 2

2 * (y^2 - y + 1) = y + 1

2y^2 - 2y + 2 = y + 1

2y^2 - 3y + 1 = 0

y = (3 +/- sqrt(9 - 8)) / (2 * 2)

y = (3 +/- 1) / 4

y = 4/4 , 2/4

y = 1 , 1/2

x^2 - 2 = 2y^3

x^2 - 2 = 2 * 1

x^2 - 2 = 2

x^2 = 4

x = -2 , 2

(-2 , 1) and (2 , 1) are the solutions

1st solution: x=2, y=1

2nd solution: x=156, y=23

If you consider negative numbers as well,

3rd solution: x=-2, y=1

4th solution: x=-156, y=23

get an excel spreadsheet make the equation in the form y= f(x), then put in integers for x and see how many you get for the f(x). Obvious is 2,1 but I don't know any other way myself. Good luck.