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Let other side be x

h² + (h - 2)² = x²

With the other leg (o) m-n = 2 so m-o = the greatest square of any odd number such that the square is less than m. 

5-3 = 2 and 5-o = 1^2, so o = 4 = √(2(5+3))

17-15 = 2 and 17-o = 3^2, so o = 8 = √(2(17+15))

37-35 = 2 and 37-o = 5^2, so o = 12 = √(2(37+35))

65-63 = 2 and 65-o = 7^2, so o = 16 = √(2(65+63))

101-99 = 2 and 101-o = 9^2, so o = 20 = √(2(101+99))

etc.

The square of the other side (o) is 2(m+n)

Let other side be t.

t^2 = m^2-n^2 = (m-n)(m+n) = 2(m+n) = 2(m-n+2n) = 4(1+n).

t =2√(1+n).

Or, t =  2√(m-1). for real values m>0.

a^2 = 4n + 4

a^2 = 4 * (n + 1)

a = 2 * sqrt(n + 1)

3 , 4 , 5 =>> 3 , 4 , 5 and 6 , 8 , 10

5 , 12 , 13 =>> 10 , 24 , 26

7 , 24 , 25 =>> 14 , 48 , 50

9 , 40 , 41 =>> 18 , 80 , 82

11 , 60 , 61 =>> 22 , 120 , 122

and so on. They work. There's no unique solution they all work

if ( m - n ) = 2

then square of the other side is m² - n²

∴

m² - n²

= ( m - n ) ( m + n )

= 2( m - n )

m = n + 2

a^2 + n^2 = m^2

a^2 + n^2 = (n + 2)^2

a^2 + n^2 = n^2 + 4n + 4

a^2 = 4n + 4

a^2 = 4 * (n + 1)

a = 2 * sqrt(n + 1)

The square of the other side is 4n + 4.  If you want integer solutions, then we can search.

3 , 4 , 5 =>>  3 , 4 , 5 and 6 , 8 , 10

5 , 12 , 13 =>> 10 , 24 , 26

7 , 24 , 25 =>> 14 , 48 , 50

9 , 40 , 41 =>> 18 , 80 , 82

11 , 60 , 61 =>> 22 , 120 , 122

and so on.  They work.  There's no unique solution.