A computer search shows that it is true at least for a and b values up to 10,000.  Further, a and b are always either both 0 mod 4 or 2 mod 4 so the sum

of a and b is always 0 mod 4.

Proof:

Squares of even integers are all 0 mod 4 since (2x)^2 = 4x^2.

Squares of odd  integers are all 1 mod 4 since (2x+1)^2 = 4x^2 + 4x + 1.

(a^2+b^2+1) mod 4 can only be {1,2,3} but (c^2) mod 4 can only be {0,1}

Thus a and b must each be even so that a^2 and b^2 can each be 0 mod 4.

So ⌊a/2⌋ and ⌊b/2⌋ are in fact a/2 and b/2.  So (a+b)/2 or the average of a and b is the quantity in question.

wlog let a <= b

Define m = (a+b)/2   and   d = (b-a)/2

Since a and b are both positive and even, then m and d are both positive integers.

Rewrite a^2+b^2+1=c^2  as:  2(m^2 + d^2) + 1 = c^2

But using the earlier mod 4 argument, m and d must each be even.

If m is even, then m = 2n, where n is an integer, 

Rewrite m = (a+b)/2 as 2n = (a+b)/2, 

thus 2n = (a+b)/2 = ⌊a/2⌋+ ⌊b/2⌋ is even.