Question

Find the Pythagorean Triplets for the number 17

Answer 1

Answer:

8,15,17

Step-by-step explanation:

We know that all solutions of the Diophantine equation x^2 + y^2 = z^2, are given by the set of the equations:

x = k(a^2 - b^2) … (1)

y = 2kab … (2)

z = k(a^2 + b^2) … (3)

with a > b

We observe that 17 ≡ 1mod4. This means that 17 can be expressed as sum of two squares, so since 17 is prime number, it follows that k = 1 and the only possibilities for 17 to be a member of a primitive Pythagorean triple, is to be of the form x = a^2 - b^2 or z = a^2 + b^2. Therefore, any pair of (a, b) forming this Pythagorean triple, will be solution of the Diophantine equations a^2 - b^2 = 17 and a^2 + b^2 = 17.

In order to solve the Diophantine equation a^2 - b^2 = 17, we have:

a^2 - b^2 = 17 => (a-b)(a+b) = 17

Now, since a > b, it follows that a+b > a-b. Therefore, we obtain:

a-b = 1 and a+b = 17 => 2a = 18 and 2b = 16, from which we take:

a = 9 and b = 8.

Therefore, the only positive solution of the Diophantine equation a^2 - b^2 = 17 is:

(a = 9, b = 8) for which we take the Pythagorean triple:

x = 9^2 - 8^2 = 17

y = 2(9)(8) = 144

z = 9^2 + 8^2 = 145

The only positive solution of the Diophantine equation a^2 + b^2 = 17 is:

(a = 4, b = 1) for which we take the Pythagorean triple:

x = 4^2 - 1^2 = 15

y = 2(4)(1) = 8

z = 4^2 + 1^2 = 17

Therefore, the only Pythagorean triples containing in one of its members the number 17, are the following two:

(8, 15, 17) and (17, 144, 145).

Answer 2

Step-by-step explanation:

$x7 \frac{1}{71} triplekv \: \: ghevb \ \sqrt[17]{29} 96 \\ .295 + 2.61 = 32.$