Question

已知x和y是正整数,且满足xy+x+y=71,x^2+xy^2=880求x^2+y^2的值

Answer 1

x^2y+xy^2=xy(x+y)=880

xy+x+y=xy+(x+y)=71

设xy=a,x+y=b

∴ab=880,a+b=71

解得：a=16,b=55或a=55,b=16

当a=16,b=55时,x、y不为正整数

∴xy=55,x+y=16

(x+y)^2=256=x^2+2xy+y^2

∴x^2+y^2=x^2+2xy+y^2-2xy=256-110=146