Solving the system { 18 x y 2 + x 3 = 12 27 x 2 y + 54 y 3 = 38 and I wonder whether there is some slick method to find the only real root ( x , y ) = ( 2 , 1 / 3 ) without relying on Cardano's formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I'm tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

Question

Solving the system      {                            18          x                      y            2                    +                      x            3                    =          12                                      27                      x            2                    y          +          54                      y            3                    =          38                        and I wonder whether there is some slick method to find the only real root   (  x  ,  y  )  =  (  2  ,  1  /  3  ) without relying on Cardano&#039;s formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I&#039;m tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

Quinn Everett · Accepted Answer

It makes sense to change variables, letting   v  =  3  y and   x  =  2  u, so that what we&#039;re looking for is   (  u  ,  v  )  =  (  1  ,  1  ). The equations become      {                            u                      v            2                    +          2                      u            3                    =          3                                      18                      u            2                    v          +                      v            3                    =          19                        Multiplying the first by   19 and the second by   3 and subtracting gives  38      u    3    −  54      u    2    +  19  u      v    2    −  3      v    3    =  (  u  −  v  )  (  38      u    2    −  16  u  v  +  3      v    2    )  =  0The quadratic term is easily seen to be non-negative:  9      v    2    −  48  u  v  +  114      u    2    =  (  3  v  −  8  u      )    2    +  50      u    2  Hence we must have   u  =  v. Plugging this into either equation gives the desired result,   u  =  v  =  1.Just to elaborate, the given equations change into something of the form      {                            r                      u            3                    +          (          1          −          r          )          u                      v            2                    =          1                                      s                      v            3                    +          (          1          −          s          )                      u            2                    v          =          1                        This leads to  (  u  −  v  )      (    r          u      2        −    (    1    −    r    −    s    )    u    v    +    s          v      2        )    =  0and the quadratic contributes nothing to the (real) solution if   (  1  −  r  −  s      )    2    &amp;lt;  4  r  s

anginih86 · Accepted Answer

Indeed, all rational solutions are given by   (  x  ,  y  )  =  (  2  ,  1  /  3  ). We can see this without any theory on elliptic curves, but just with a computation: Substitute       y    2    =  (  12  −      x    3    )  /  (  18  x  ) from the first equation into the second to obtain   y  =  19  x  /  (  6  (  2      x    3    +  3  ). This in turn gives the following result:  (  x  −  2  )  (      x    2    +  2  x  +  4  )  (  2      x    2    +  4  x  +  3  )  (  4      x    4    −  8      x    3    +  10      x    2    −  12  x  +  9  )  =  0Obviously, only   x  =  2 is a rational solution for   x. Also we see, that it is the only real solution.

Solving the system { <mtable columnalign="left left" rowspacing=".2em" columnspacing

Answered question

Answer & Explanation

New Questions in Algebra I