Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5, −3, and −1. Find three other consecutive odd integers that also satisfy the given conditions.

What are the integers?

Question

Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5​, −3​, and −1. Find three other consecutive odd integers that also satisfy the given conditions.

What are the​ integers?

rock_mit182 · Answer

$$X_1 = (2n+1) ; X_2 = (2n+1) +2; X_3 = (2n+1) +4$$
$$(X_3)^2 = (X_2 +X_1)^2 -33$$

rock_mit182 · Answer

anyone could check my work ?

phi · Answer

I think
 sum of the squares of the first two
means $x_1^2 + x_2^2 $

rock_mit182 · Answer

oh i see could you help with a hard problem ?

rock_mit182 · Answer

cause i dont know hot translate it into a math equation ..

rock_mit182 · Answer

how*

rock_mit182 · Answer

A 28​-ft by 42​-ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 456 ft2​, how wide is the​ walkway?
The width of the walkway is 
_________ feet.

rock_mit182 · Answer

i dont have a clue please help me out

phi · Answer

can you make that a new post for that question ?

phi · Answer

for the 3 odd integers, and using a,b, and c

a^2 + b^2 = c^2+33

let a= b-2,   and c= b+2

(I'm ignoring if b is even or odd... I expect to find an odd answer!)

(b-2)^2 + b^2 = (b+2)^2 + 33

expand, simplify, and get
$$ b^2 -8b -33=0$$
solve for b

rock_mit182 · Answer

ok thank you so much