Diophantine equations and number theory

Question

If $x$, $y$ and $z$ are integers, then prove that $$x^2+y^2-z^2=1997$$ has infinitely many integer solutions.

Answer 1

If you begin with a solution with $x,y,z > 0,$ you can get solutions with larger and larger entries from $$ (x,y,z) \mapsto (2x+y+2z, \; \; x + 2y + 2z, \; \; 2x+2y+3z) $$

jagy@phobeusjunior:~$ ./homothety_indef 1 1 -1 0 0 0  1 1 -1 0 0 0  3


Sat Mar 18 12:21:06 PDT 2017

====================================...............

      2     -1      2 transposed        2      1      2
      1     -2      2 transposed       -1     -2     -2
      2     -2      3 transposed        2      2      3

      2      1     -2 transposed        2      1      2
      1      2     -2 transposed        1      2      2
      2      2     -3 transposed       -2     -2     -3

      2      1      2 transposed        2      1      2
      1      2      2 transposed        1      2      2
      2      2      3 transposed        2      2      3



             -4 :     1     1         -1      0    0    0
             -4 :     1     1         -1      0    0    0


-4 =  -1 * 2^2
-4 =  -1 * 2^2
Sat Mar 18 12:21:06 PDT 2017


jagy@phobeusjunior:~$