Given $x,y,s$ are natural numbers:
$$x^3-2xy^2+y^3-s^2=0$$
I found the solutions using wolfram alpha
$$(x,y,s) = (1,2,1), (6,10,4), (4,8,8)$$
But how do I prove these are the only solutions? Any tips or reference to papers that study this diophantine equation would be much appreciated.
On
Here are a number of solutions with cube-free $s.$ For any of these, we get other solutions with $(x t^2, y t^2, s t^3).$ There are so very many because $$ x^3 - 2 x y^2 + y^3 = (x-y)(x^2 + xy - y^2). $$ Furthermore, if $\gcd(x,y) = 1, $ then $\gcd(x-y, x^2 + xy - y^2)=1.$
I don't see any reason to think that all solutions can be described.
x: 1 = 1 y: 2 = 2 s: 1 = 1
x: 6 = 2 3 y: 10 = 2 5 s: 4 = 2^2
x: 10 = 2 5 y: 5 = 5 s: 25 = 5^2
x: 10 = 2 5 y: 30 = 2 3 5 s: 100 = 2^2 5^2
x: 17 = 17 y: 8 = 2^3 s: 57 = 3 19
x: 19 = 19 y: 95 = 5 19 s: 722 = 2 19^2
x: 24 = 2^3 3 y: 193 = 193 s: 2327 = 13 179
x: 30 = 2 3 5 y: 21 = 3 7 s: 99 = 3^2 11
x: 30 = 2 3 5 y: 130 = 2 5 13 s: 1100 = 2^2 5^2 11
x: 33 = 3 11 y: 22 = 2 11 s: 121 = 11^2
x: 33 = 3 11 y: 58 = 2 29 s: 95 = 5 19
x: 33 = 3 11 y: 132 = 2^2 3 11 s: 1089 = 3^2 11^2
x: 40 = 2^3 5 y: 65 = 5 13 s: 25 = 5^2
x: 55 = 5 11 y: 99 = 3^2 11 s: 242 = 2 11^2
x: 57 = 3 19 y: 133 = 7 19 s: 722 = 2 19^2
x: 60 = 2^2 3 5 y: 105 = 3 5 7 s: 225 = 3^2 5^2
x: 66 = 2 3 11 y: 22 = 2 11 s: 484 = 2^2 11^2
x: 66 = 2 3 11 y: 165 = 3 5 11 s: 1089 = 3^2 11^2
x: 70 = 2 5 7 y: 21 = 3 7 s: 539 = 7^2 11
x: 70 = 2 5 7 y: 170 = 2 5 17 s: 1100 = 2^2 5^2 11
x: 73 = 73 y: 57 = 3 19 s: 316 = 2^2 79
x: 76 = 2^2 19 y: 57 = 3 19 s: 361 = 19^2
x: 89 = 89 y: 890 = 2 5 89 s: 23763 = 3 89^2
x: 97 = 97 y: 88 = 2^3 11 s: 303 = 3 101
x: 102 = 2 3 17 y: 778 = 2 389 s: 18668 = 2^2 13 359
x: 115 = 5 23 y: 95 = 5 19 s: 550 = 2 5^2 11
x: 126 = 2 3^2 7 y: 226 = 2 113 s: 820 = 2^2 5 41
x: 136 = 2^3 17 y: 425 = 5^2 17 s: 5491 = 17^2 19
x: 145 = 5 29 y: 29 = 29 s: 1682 = 2 29^2
x: 145 = 5 29 y: 116 = 2^2 29 s: 841 = 29^2
x: 145 = 5 29 y: 120 = 2^3 3 5 s: 775 = 5^2 31
x: 145 = 5 29 y: 314 = 2 157 s: 2327 = 13 179
x: 145 = 5 29 y: 870 = 2 3 5 29 s: 21025 = 5^2 29^2
x: 145 = 5 29 y: 986 = 2 17 29 s: 26071 = 29^2 31
x: 153 = 3^2 17 y: 442 = 2 13 17 s: 5491 = 17^2 19
x: 154 = 2 7 11 y: 253 = 11 23 s: 363 = 3 11^2
x: 195 = 3 5 13 y: 51 = 3 17 s: 2556 = 2^2 3^2 71
x: 208 = 2^4 13 y: 377 = 13 29 s: 1859 = 11 13^2
x: 210 = 2 3 5 7 y: 205 = 5 41 s: 475 = 5^2 19
x: 228 = 2^2 3 19 y: 57 = 3 19 s: 3249 = 3^2 19^2
x: 246 = 2 3 41 y: 205 = 5 41 s: 1681 = 41^2
x: 264 = 2^3 3 11 y: 433 = 433 s: 767 = 13 59
x: 273 = 3 7 13 y: 442 = 2 13 17 s: 169 = 13^2
x: 286 = 2 11 13 y: 165 = 3 5 11 s: 3509 = 11^2 29
x: 286 = 2 11 13 y: 962 = 2 13 37 s: 19604 = 2^2 13^2 29
x: 290 = 2 5 29 y: 470 = 2 5 47 s: 300 = 2^2 3 5^2
x: 304 = 2^4 19 y: 593 = 593 s: 4777 = 17 281
x: 310 = 2 5 31 y: 186 = 2 3 31 s: 3844 = 2^2 31^2
x: 315 = 3^2 5 7 y: 295 = 5 59 s: 1450 = 2 5^2 29
x: 349 = 349 y: 205 = 5 41 s: 4668 = 2^2 3 389
x: 375 = 3 5^3 y: 231 = 3 7 11 s: 5004 = 2^2 3^2 139
x: 377 = 13 29 y: 638 = 2 11 29 s: 2523 = 3 29^2
x: 385 = 5 7 11 y: 330 = 2 3 5 11 s: 3025 = 5^2 11^2
x: 385 = 5 7 11 y: 630 = 2 3^2 5 7 s: 1225 = 5^2 7^2
x: 385 = 5 7 11 y: 660 = 2^2 3 5 11 s: 3025 = 5^2 11^2
x: 390 = 2 3 5 13 y: 165 = 3 5 11 s: 6525 = 3^2 5^2 29
x: 406 = 2 7 29 y: 357 = 3 7 17 s: 2989 = 7^2 61
x: 408 = 2^3 3 17 y: 697 = 17 41 s: 3179 = 11 17^2
x: 427 = 7 61 y: 183 = 3 61 s: 7442 = 2 61^2
x: 465 = 3 5 31 y: 186 = 2 3 31 s: 8649 = 3^2 31^2
x: 465 = 3 5 31 y: 754 = 2 13 29 s: 697 = 17 41
x: 481 = 13 37 y: 40 = 2^3 5 s: 10479 = 3 7 499
x: 481 = 13 37 y: 312 = 2^3 3 13 s: 6929 = 13^2 41
x: 505 = 5 101 y: 105 = 3 5 7 s: 10900 = 2^2 5^2 109
x: 568 = 2^3 71 y: 497 = 7 71 s: 5041 = 71^2
x: 574 = 2 7 41 y: 133 = 7 19 s: 13083 = 3 7^2 89
x: 577 = 577 y: 528 = 2^4 3 11 s: 4193 = 7 599
There is a beautiful parameterization by Lagrange and Legendre:
$$x^3 + a x^2y + b x y^2 + c y^3 = z^2\tag1$$
where,
$$x = u^4 - 2b u^2 v^2 - 8c u v^3 + (b^2 - 4a c)v^4$$
$$y = 4v(u^3 + a u^2v + b u v^2 + c v^3)$$
and $z$ is a $6$-deg polynomial.
In your case, with $a,b,c = 0,-2,1$, we have,
$$x^3 - 2x y^2 + y^3 = z^2$$
where,
$$x = u^4 + 4 u^2 v^2 - 8 u v^3 + 4 v^4$$
$$y = 4 v (u^3 - 2 u v^2 + v^3)$$
$$z = u(u^5 - 10 u^3 v^2 + 20 u^2 v^3 - 20 u v^4 + 8 v^5)$$
An infinite set of solutions is found when $y=2x$. In this case the equation $x^3 - 2xy^2 + y^3 = s^2$ reduces to: $$x^3 = s^2$$ So we can choose $x=a^2$ for any positive integer $a$, so then $a^6 = s^2$, so $s=a^3$. So the following combination works for any positive integer $a$: $$(x,y,s) = (a^2, 2a^2, a^3)$$