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Three consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 3, 5, and 7. Find three other consecutive odd integers that also satisfy the given conditions.

Respuesta :

tonb

Answer:

1, 3 and 5

Step-by-step explanation:

You can formulate the postulate as:

x² + (x + 2)² + 15 = (x + 4)²

Which simplifies to x² - 4x + 3 = 0, with solutions x=1 and x=3

so 1, 3 and 5 must be the other tuple.

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