Question

15. Among four consecutive odd numbers the sum of the
squares of the first two numbers is 64 less than the
sum of the squares of the next two numbers. Find the
four numbers
(1) 7, 9, 11, 13
(2) 3, 5, 7, 9
(3) 5, 7, 9, 11
(4) 1, 3, 5, 7​

Answer 1

The four consecutive odd numbers are 1, 3, 5, 7

Solution:

Let the four consecutive odd numbers be:

x , x + 2, x + 4, x + 6

The sum of the  squares of the first two numbers is 64 less than the  sum of the squares of the next two numbers

According to this condition,

$x^2 + (x+2)^2 = [(x+4)^2 + (x + 6)^2] - 64\\\\x^2 + x^2 + 4x + 4 = [x^2 + 16+8x + x^2 + 36+12x] - 64\\\\2x^2 + 4x + 4 = 2x^2 +20x +52-64\\\\2x^2 + 4x + 4 = 2x^2 +20x -12\\\\4x + 4 = 20x - 12\\\\20x - 4x = 4+12\\\\16x = 16\\\\x = 1$

Therefore, four numbers are:

x  = 1

x + 2 = 1 + 2 = 3

x + 4 = 1 + 4 = 5

x + 6 = 1 + 6 = 7

Thus the four consecutive odd numbers are 1, 3, 5, 7

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