若x1,x2为正整数,且x1+x2=a,x1x2=1-b,求证：a的平方+b的平方为合数.

x1+x2=a,x1x2=1-b,代入a^2+b^2,得到a^2+b^2=(x1+x2)^2+(1-x1x2)^2=x1^2+2x1x2+x2^2+1-2x1x2+(x1x2)^2=x1^2+x2^2+(x1x2)^2+1=（x1^2+1）(x2^2+1)∵x1,x2为正整数∴（x1^2+1）≠1,(x2^2+1)≠1有两个不等于1的因子.所以...