设a、b、C均为奇数,原题成立,不妨设a=2m+1,b=2n+1,c=2p+1（m、n、p均为整数）则a²+b²=(2m+1)²+(2n+1)²=4m²+4m+1+4n²+4n+1=4(m²+m+n²+n)+2因4(m²+m+n²+n)为偶数,...