Is the solution just 3600,2500 and 0036,0025.
Reasoning: Since there needs be a difference of 1s and we know that M,N are squares. a^2-b^2=(a+b)(a-b) and the differences must be one of {1100,1010,1001,0110,0101,0011}, and u can show quite quickly that for all other than 1100 and 0011, there exists no solutions. leaving the answers 3600,2500 and 0036,0025. To show that they don't satify, break down the difference into prime factorisation and there'll be 2C(number of factors) possible combinations, and u can eliminant most since odd(O)+/-even(E)=O, O+/-O=E, E+/-E=E. Then just add the numbers (a+b),(a-b) together and half it to get 'a', then square 'a' and see whether it has two of the same numbers in the places of the '00', for most cases, this doesn't match, so its not a possible pairing, so eliminate it, then we get left with answer pairs of 3600,2500 and 0036,0025. If 4-digit integers counts for something that begins with 0, then keep 0036,0025; else there's only one pair.