Sum of Squares: Special Case

A special case for the sum of squares exists with specific conditions. Arrange two numbers, $ab$ and $cd$, such that:

• The unit's digit of the first number ($b$) is one greater than the ten's digit of the second number ($c$).
• The sum of the ten's digit of the first number ($a$) and the unit's digit of the second number ($d$) is 10.

If these conditions are met, the answer is the sum of the squares of the digits of the first number, multiplied by 101. That is, $(a^2 + b^2) \times 101$. For example, for $72^2 + 13^2$, the conditions are met (2 = 1+1 and 7+3=10). The answer is $(7^2 + 2^2) \times 101 = 53 \times 101 = 5353$.