The sum of the squares of the largest and smallest of three consecutive odd integers is 353 less than 3 times the square of the middle one. Find the integers.

three consecutive odds:

x-2 , x, x+2

(x-2)^2 + (x+2)^ = 3x^2 - 353
x^2 - 4x + 4 + x^2 + 4x + 4 = 3x^2 - 353
x^2 = 361
x = ±√361
x = ± 19

So the integers are:
17, 19, and 21
or
-17, -19, -21

Find three consecutive integers such that the product of the two largest is 29 more than the square of the smallest integer. (Enter your answers as a comma-separated list.)

Let's represent the three consecutive odd integers as (2n-1), (2n+1), and (2n+3), where n is an integer.

According to the problem statement, the sum of the squares of the largest and smallest integers is 353 less than 3 times the square of the middle integer:

[(2n-1)^2 + (2n+3)^2] = 3(2n+1)^2 - 353

Expanding this equation:
(4n^2 - 4n + 1 + 4n^2 + 12n + 9) = 12n^2 + 12n + 3 - 353

Combining like terms:
8n^2 + 8n + 10 = 12n^2 + 12n - 350

Rearranging the terms and simplifying:
4n^2 + 4n - 360 = 0

Dividing the equation by 4:
n^2 + n - 90 = 0

Factoring the quadratic equation, (n + 10)(n - 9) = 0:
n + 10 = 0 or n - 9 = 0

Solving for n:
n = -10 or n = 9

Therefore, the two sets of consecutive odd integers are:
1) (2n-1), (2n+1), (2n+3) = (-19, -17, -15)
2) (2n-1), (2n+1), (2n+3) = (17, 19, 21)

So, the consecutive odd integers are either (-19, -17, -15) or (17, 19, 21).

To solve this question, let's assume the three consecutive odd integers as (n-2), n, and (n+2), where n represents the middle integer.

We are given that the sum of the squares of the largest and smallest of the three integers is 353 less than 3 times the square of the middle one. So, we can write the equation:

(n+2)^2 + (n-2)^2 = 3n^2 - 353

Expanding the equation:

n^2 + 4n + 4 + n^2 - 4n + 4 = 3n^2 - 353

Simplifying and combining like terms:

2n^2 + 8 = 3n^2 - 353

Rearranging the equation:

3n^2 - 2n^2 - 361 = 0

n^2 - 361 = 0

To solve this equation, we can factor it as:

(n + 19)(n - 19) = 0

Setting each factor equal to zero:

n + 19 = 0 or n - 19 = 0

Solving for n:

n = -19 or n = 19

Since we are looking for odd consecutive integers, the middle integer cannot be -19. Therefore, n = 19.

Substituting the value of n back into our assumption, the three consecutive odd integers are:

(n-2) = 19 - 2 = 17
n = 19
(n+2) = 19 + 2 = 21

So, the three consecutive odd integers are 17, 19, and 21.