The difference between two positive integers is 56.
If their squares have both the same units digit and the same tens digit, find the sum of the two positive integers.

1. Find the minimal couple, complying with the above, and the corresponding sum.
2. Define formulas for additional solutions (both numbers and their sums).

The pair {78,22} have a difference of 56, while their squares are 6084 and 484. 78+22=100. Obviously, if squares have the same two terminal digits, their difference will be divisible by 100.

a=(78+25k), b=(22+25k) gives a^2-b^2=2800(k+2). The sum in each case is 50(k+2), for all k.

Edited on February 29, 2016, 8:27 am

Posted by broll on 2016-02-29 08:25:26