Conference rusure::math

521.0. "solving n unknowns in one equation" by RDGE28::FLASH (project account) Wed Jun 25 1986 09:12

    (Simon Clinch...)				 A  A	  A
    						  1  2     n
    Given that P is a sum of terms of the form kx  x  ...x  
    						 1  2     n

    with each k a real constant and x  all real,  and A  integers >= 0,
		                     i                 i

    				    n
    (in short P is a polynomial in P ),
    
    prove that there exists such a P so that the single equation:
    
		P(x ,x ,...,x ) = 0
    		   1  2      n

    can be solved for all x .
    			   i

    (This is easier than it may seem -- use lateral thinking!)

    How about
      2*j1   2*j2   2*j3
    x1    +x2    +x3    +... = 0
    
    which has the solution(s) Xi = 0, i=1..n for arbitrary integers
    Ji>0.