solve this?

Question

there are four odd consecutive integers. if the first term added to twice the quanity of the sum of the third and fourth term is equal to the second term squared minus three, find these numbers.

Answer 1

let n the first, the others are n + 2, n + 4, n + 6

we have the equation :

n + 2(n + 4 + n + 6) = (n + 2)² - 3

5n + 20 = n² + 4n + 1

n² - n - 19 = 0 : no solution with n as integer ? !!!

What's the problem ?


Additional details for Marbledo :

(n + 2)² - 3 = n² + 4n + 4 - 3 = n² + 4n + 1

and Marbledo has

(x + 2)² - 5x - 23 = 0
or x² + 4x + 4 - 5x -23 = x² -x -19 = 0

it's the same equation as mine ! No ?

Answer 2

@Nestor
(n + 2)² - 3 cannot be simplified. The exponent does not distribute.

Let the first number = x. Our four consecutive odd numbers are x, x+2, x+4, and x+6.

The first term [x] added to twice the quanity of the sum of the third and fourth term [2(x+4 + x+6)] is equal to the second term squared minus three [(x+2)² - 3].

x + 2(x+4 + x+6) = (x+2)² - 3;
x + 2(2x + 10) = (x+2)² - 3;
x + 4x + 20 = (x+2)² - 3;
5x + 23 = (x+2)²;
0 = (x+2)² - 5x - 23;

This quadratic does not produce an integer value for x. I don't think there is an answer to your question. Can anyone correct me on this?