Question

Find a Pythagorean triplet whose greatest member is 101.​

Answer 1

The number 101 is the 26th prime and the first prime after 100. Since the number 101 is prime and the greatest member of a Pythagorean triple, then it must belong to a primitive pythagorean triple. This is important since 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)

Here, we need to have z = 101, therefore k = 1 and the triple is primitive. This means that:

z = 101 => a^2 + b^2 = 101 … (4)

Now, since 101 is prime and since 101 ≡ 1mod4, it follows that 101 can be expressed as sum of two squares, since it does not contain a prime factor congruent to 3mod4, raised to an odd power.

The only positive integral solutions of (4) are:

(a = 1, b = 10), (a = 10, b = 1)

From these the only acceptable is (a = 10, b = 1). This solution gives the unique Pythagorean triple satisfying the condition of our problem, which is:

x = 10^1 - 1^2 = 99

y = 2(10)(1) = 20

z = 10^ + 1^2 = 101

Answer 2

Step-by-step explanation:

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