Three consecutive odd integers such that four times the middle integer is two more than the sum of the first and third

Bearing in mind that, if you multiply, any integer whatsoever by 2, you end up with a even integer, 17*2, or 19*2, 88*2 you name it, you get an even number

now, you can get an odd integer by simply going from an even integer, back or forth, so if you have an even number of say 24, 24 - 1, 23, 24+1, 25, 23 and 25 are the odd ones next to 24

so ... let's pick some number, let's say hmm "a", we know 2*a is even, so 2a, thus, 2a + 1 or 2a - 1 is an odd one ... let's use hmm 2a + 1 as our first odd integer

to jump from an odd integer to another, you simply add 2, 3+2, 5, 5+2, 7 and so on

so ... our first one is 2a + 1, so the next consecutive one, will then be (2a+1) + 2

and next consecutive after that is (2a+1+2) + 2

so our three consecutive odd integers are then

2a + 1

2a + 3

2a + 5

now, 4 times the middle is 4 (2a+3)

the sum of the first and last is (2a+1) + (2a+5)

now, two more than that is just (2a+1) + (2a+5) + 2

thus

4 (2a + 3) = (2a + 1) + (2a + 5) + 2

solve for "a"