Solve the system in natural numbers.

[Math Tutor]

[tex]\begin{array}{|l} x+y-z=6 \\ x^{2}+y^{2}-z^{2}=6 \end{array}[/tex]

Solve the system in natural numbers.

[Guest]

4 pairs of solutions in natural numbers (x,y,z)

YS.

[Guest]

hehe - Nothing new around ????

[Math Tutor]

Yes, it seems that noone can solve the system.

[Guest]

This is a good joke. Well, I was thinking that there is another reason in place for that this problem
is still up. Now I understand You doubted my words and I need to type the whole solution here. I just
wonder if I was wrong when i gave the solutions for the previous problems. As far as i know, all were
perfect. Anyway, let's solve it the way you want (the long way):

z = x + y -6 (making this replacement in the second equation)
x^2 + y^2 - (x + y - 6)^2 = 6
12 x - 2 x y + 12 y - 42 = 0
2x(6 - y) + 12 y - 42 - 30 = - 30
2x(6 - y) - (72 - 12y) = - 30
2x(6 - y) - 12(6 - y) = - 30
2(x - 6)(6 - y) = - 30
(6 - x)(6 - y) = 15

The only fitting solutions in accordance with the problem requirements are: (7, 21); (21, 7); (9, 11); (11, 9).
The full solutions (x,y,z): (7, 21, 22); (21, 7, 22); (9, 11, 14); (11, 9, 14).