Question

find the Pythagorean triplet whose largest member is 65

Answer 1

HEYA!
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✴PYTHAGOREAN TRIPLET ✴
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$= > a \: pythagorean \: triplet \: is \: represented \: by \: 2m \: \: m {}^{2} + 1 \: \: and \: m {}^{2} - 1$
$hence \: the \: largest \: member \: is \: m {}^{2} + 1$
$= > m {}^{2} + 1 = 65 \\ \\ = m {}^{2} = 64 \\ \\ = > m = 8 \\ \\ = > 2m = 8 \times 2 \\ \\ = 16 \\ \\ m {}^{2} - 1 = 64 - 1 = 63$
✨Hence the triplet is 65 , 16 and 63

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Answer 2

The required Pythagorean triplet is 16, 63, 65.

m, n, p form a Pythagorean triplet if,

m² + n² = p²

2m, m² - 1, m² + 1 make a Pythagorean triplet.

⇒(2m)² + (m²-1)²

⇒ 4m² + (m²)² + 1 - 2m²

⇒ (m²)² + (1)² + 2(m²)(1)

⇒ (m² + 1)²

So, 2m, m² - 1, m² + 1 form a Pythagorean triplet where m² + 1 is the largest number.

Given,

Largest number is 65

⇒ m² + 1 = 65

⇒m² = 65 - 1

⇒ m² = 64

⇒ m = 8 ( Considering only positive value).

Now

Pythagorean triplet with m = 8,

⇒ 2m = 2(8) = 16

⇒m² - 1 = 8² - 1 = 63

⇒m² + 1 = 8² + 1 = 65

Therefore, 16, 63, 65 is a Pythagorean triplet with largest number as 65.