This is one of my favorite NS shortcuts because the algebra is so cool...when this question first came to my attention (I used to say when it was first put on a NS test, but I quickly learned that was almost always an inaccurate statement to use concerning NS...what goes around will most certainly eventually come back around...), we had to wait 3 long months for it to reappear on the 1st test of the year after being on the State test the previous year...not enough of a pattern to understand what was happening...we spent the summer guessing (mostly incorrectly)...of course, that is the fun part, but I digress...

Let this problem be modeled for a second by (AB)^2 + (CD)^2, where A, B, C & D represent digits (for example, in 36^2 + 57^2, A=3, B=6, C=5 & D=7)

To fit the structure of this problem, A+D=10 and B=C+1 (B and C are descending).

Now, to describe the algebraic model for this problem, let's refer the the standard model of a 2-digit number, 10x+y...using 10x+y for the 1st number (AB), the entire problem would be set up like this:

 (10x+y)^2 + [ 10(y-1) + (10-x)]^2 =

100x^2 + 20xy + y^2 + [10y-10 +10-x]^2 =

100x^2 + 20xy + y^2 + 100y^2 - 20xy +x^2 =

101x^2 + 101y^2 = 101( x^2 + y^2), where x=A & Y=B

So, 36^2 + 57^2 = 101(3^2 + 6^2) = 101(9+36)= 4545..the 2nd number, 57,  in the problem becomes inmaterial to the calculation process.  

As for #57 on Test 2, 54^2 + 66^2. To get it in the proper form, switch the numbers being squared, which means that 101(6^2 +6^2) = 7272

Boy, that was sure cooler to do on a chalkboard in front of a bunch of "oohing and aahhing" NS players than "picking and grinning" alone behind the keyboard...hopefully it made sense because it truly is one of my favorite tricks...(nothing like diving right in the deep end with your first posting, huh???...)