(2)当k=3时,若存在正整数m,n,满足m(m+3)=n(n+1),则4m2+12m=4n2+4n,(2m+3)2=(2n+1)2+8,(2m+3-2n-1)(2m+3+2n+1)=8,(m-n+1)(m+n+2)=2,而m+n+2>2,故上式不可能成立. (10分)
当k≥4时,若k=2t(t是不小于2的整数)为偶数,取m=t2-t,n=t2-1则m(m+k)=(t2-t)(t2+t)=t4-t2,
n(n+1)=(t2-1)t2=t4-t2,因此这样的(m,n)满足条件.若k=2t+1(t是不小于2的整数)为奇数,取
m=
t2−t |
2 |
t2+t−2 |
2 |
t2−t |
2 |
t2−t |
2 |
1 |
4 |
t2+t−2 |
2 |
t2+t |
2 |
1 |
4 |
(15分)