Problema Solution
Three consecutive odd integers are such that the square of the third integer is seven greater than the sum of the squares of the first two. one solution is -1, 1, and 3. Find the three other consecutive odd integers that also satisfy the given conditions.What are the integers?
Answer provided by our tutors
let '2n+1' represent the first odd integer, then '2n+3' and '2n+5' represent the second and third consecutive odd integers
(2n+5)^2 = 7 + ((2n+1)^2 + (2n+3)^2)
solving for 'n' we have n={2, -1}
2*2 + 1 = 5
2*(-1) + 1 = -1
the first odd integer could also be 5, giving, 5, 7 and 9 as the second set of such integers