Solving x2 + y2 with xy+x+y=71 and xy(x+y)=880

[murshid_islam]

i have to find x² + y² if x and y are positive integers and
xy + x + y = 71
xy(x + y) = 880

this is what i have done:
since x and y divides 880, x or y must be 2, 5 or 11, because 880 = 2⁴.5.11

from the first equation i can see that
x = 11 and y = 5 or x = 5 and y = 11

so x² + y² = 146

but is there a way to do it algebraically?

thanks in advance.

 

[dmoravec]

one thing you could do is from equation 1 set x+y=71-xy, plug that into equation 2 to get xy(71-xy)=880. Set xy=z and you have a quadratic equation in which one solution is 55, therefore xy=55 and you get the same solution (you also get a solution of xy=16...but that doesn't fit your first equation)

 

[D H]

You can completely solve for $x^2+y^2$ algebraically, and eliminate the non-integer solutions as a final step. Make the substitutions $u=xy$, $v=x+y$. Then $x^2+y^2=v^2-2u$. With those substitutions, the two given equations become

$$u+v = 71$$

$$uv = 880$$

The two solutions to the given equations are $u=55$ and $u=16$. The goal is to solve for $x^2+y^2$. Note that
$$x^2+y^2=v^2-2u = (71-u)^2-2u = 71^2-144u+u^2$$

Applying the two solutions for $u$ yields $x^2+y^2 = 146 (u=55)$ and $x^2+y^2 = 2993 (u=16)$.

As a final step, note that the former solution, $x^2+y^2 = 146$, corresponds to $u=55,v=16$, which has integer solutions in $x$ and $y$. The solution $x^2+y^2 = 2993$ corresponds to $u=16,v=55$, which has irrational solutions in $x$ and $y$. Thus $x^2+y^2 = 146$.