Given two positive even integers, it turns out that the square of the largest minus 10 times the smaller is 76. What is the sum of the integers?

Answer 1

The answer is 26.

Here is the complete solution:

integers X and y such that 0 < y < x <--- x is the larger of the even integers

and

x = 2m and y = 2n for positive integers m and n <--- they are even integers, so they can be divided by 2

X^2 - 10Y = 76 <-- square of the largest minus 10 times smaller is 76

(2m)^2 - 10(2n) = 76 <--- substitution

4m^2 - 20n = 76 <--- squares and simplifies

m^2 - 5n = 19 <--- divides everything by 4

m^2 = 5n + 19 <--- solves for m^2

Since it must be a perfect square, the first time it happens is

for n=6, m^2 = 49 ---> m = 7

14 and 12 are the even integers

14+12=26