Find a Pythagorean triplet with 17 as one of the following members
Answers
Step-by-step explanation:
or Any natural number
m > 1,2m, m2-1
m2 + 1 forms a Pythagorean triplet
let us take m2 + 1 = 17 m2 = 16m = 4
===> value of m is a integer
so , this case is possible
so we have
2×4, 4 2 - 1 , 4 2 + 1 = 8 , 15 , 17
sorry late answer
Answer:
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-:
8^2 + 15^2 =64 + 225 = 289 = 17^2
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).
Step-by-step explanation: