Find four consecutive positive integers such that the product of the first and fourth is four less than twice the first multiplied by the fourth.

Four consecutive integers ... n, n+1, n+2, n+3.

The product of the 1st and 4th is four less than twice the 1st multiplied by the 4th.

n (n+3) = 2n (n+3) - 4 perform indicated multiplications ...

n^2+3n=2n^2+6n-4 subtract n^2 from both sides

3n=n^2+6n-4 subtract 3n from both sides

n^2+3n-4=0 factor

n^2-n+4n-4=0

n (n-1) + 4 (n-1) = 0

(n+4) (n-1) = 0, and since n>0

n=1

So the four numbers are 1, 2, 3, 4

check ...

1 (4) = 2 (1) 4-4

4=8-4

4=4