## anonymous 5 years ago the sum of the squares of two consecutive, positive even integers is 41. find the integers

1. anonymous

a miracle since if you square an even integer you get an even integer and if you add two even integers together you get another even integer.

2. anonymous

relax the "even" part and the numbers 4 and 5 work

satellite is right. Below is how you would solve the problem if you didn't specifically require even numbers: $a^2 + (a + 1)^2 = 41$ $a^2 + a^2 + 2a + 1 = 41$ $2a^2 + 2a + 1 = 41$ $2a^2 + 2a - 40 = 0$ Since $$b^2 - 4ac = 4 - (4 \cdot 2 \cdot -40) = 4 + 320 = 324 > 0$$, we know this equation has a solution. You can use the quadratic formula to find a. If you want to run it with even integers, you know that two even numbers must be two apart; you do the same as above: $a^2 + (a + 2)^2 = 41$ $a^2 + a^2 + 4a + 4 = 41$ $2a^2 + 4a + 4 = 41$ $2a^2 + 4a - 37 = 0$ But, running this through the quadratic formula will leave you with a decimal number, not an integer.