Solution: According to the known equation:

(m？ -n？ )+(m-n)=-m

(m+n)(m-n)+(m-n)=-m

(m+n+ 1)(m-n)=-m

(m+n+ 1)(n-m)=m

Since both m and n are positive integers, we know from the above formula that (n-m)≥ 1, that is, n≥m+ 1,

So: m = (m+n+1) (n-m) ≥ m+n+1,

Available: n+ 1≤0, which is obviously invalid;

Therefore, the positive integer solution satisfying m(m+2)=n(n+ 1) does not exist;

2、

Solution: (m+n+1) (n-m) = (k-1) m, ................

Because k≥3, we can get: n-m >; 0, namely: n>m, n/m > 1,

Then there is: n/m = (m+k)/(n+1) >1,

So: m+k > n+ 1，

Therefore: m

As can be seen from the above, there are k- 1 positive integers from m to m+k,

But when k=3, then: m

When n=m+ 1, we get: 2m+2=2m, which is obviously not true;

When n=m+2, we get: 2(2m+3)=2m, which is obviously not true;

But when k≥4, m

b(2m+b+ 1)=(k- 1)m

Solution: m=(b? +b)/(k- 1-2b)，

Then k- 1-2b≥ 1, and b≤(k-2)/2,

Therefore, the range of b is: 1≤b≤(k-2)/2,

If k=4, then 1≤b≤ 1, then there is

When b= 1, m=2/(3-2)=2, n=2+ 1=3,

Because there is no definite k, we can't find the general solution of this problem, but as long as k≥4, there must be positive integers m and n, which makes m(m+k)=n(n+ 1) hold.