if m>n, can 2m(m+n) be a perfect square?

Question

I'm trying to solve a problem that ask can right angle triangle have a square perimeter and it comes to the point that ask, can 2m(m+n) be a perfect square if m>n.

Answer 1

I think we can take this one step farther.

 

Let m = 8n, with both m and n positive.

 

Then

 

2m(m+n) = 2(8n)(8n + n) = 2(8n)(9n) = 144n² = (12n)², a perfect square.

 

Patrick's solution is the case where n = 1.

Answer 2

Yes. Let M=8 and N=1.  Then 2m(m+n) = 2*8*(8+1) = 2*8*9 = 144 which is a perfect square.

 

If negatives are allowed,  another example is M=8 and N= -4 which results in a perfect square of 64.