Question

find the Pythagorean triplet whose one member is 16

Answer 1

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For any natural number m > 1, we have (2m) ² + (m2 – 1)²= (m2 + 1)² . So, 2m, m² – 1 and m² + 1 forms a Pythagorean triplet.

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Let 2 m = 16,

 m = 8


 m² + 1 = 8² + 1 = 64 + 1 = 65


m²- 1 = 8 2 - 1 = 64 - 1 = 63


check:

16² + 63² = 256 + 3969 = 4225 = 65²

Answer 2

If a² + b² = c², then a = m² - n², b = 2mn, and c = m² + n²

where a, b, c are Pythagorean triplets (counting numbers) and m, n are numbers that m > n, m ≠ n, m ≥ 2 and n ≥ 1.

Here we can consider 16 as b = 2mn because both 16 and 2mn are even numbers.

2mn = 16

mn = 16 / 2 = 8

∴ m, n are factors of 8.

8 = 8 x 1

∴ We can take m as 8 and n as 1. (Not m = 1 and n = 8 as m > n.)

a = m² - n² = 8² - 1² = 64 - 1 = 63

b = 2mn = 2 x 8 x 1 = 16 (We know)

c = m² + n² = 8² + 1² = 64 + 1 = 65

∴ 63, 16 and 65 are one Pythagorean triplet.

8 = 4 x 2

∴ Consider m = 4 and n = 2.

a = 4² - 2² = 16 - 4 = 12

b = 2 x 4 x 2 = 16

c = 4² + 2² = 16 + 4 = 20

∴ 12, 16 and 20 are another Pythagorean triplet.

We can also consider 16 as a = m² - n².

16 is a multiple of 4.

Every multiples of 4 can be written as the difference of squares of two consecutive odd/even numbers. And that odd/even numbers are of next and previous numbers of 1/4 of this multiple, i. e., the average of these two consecutive odd/even numbers is 1/4 of the multiple of 4.

E. g., consider 36.

36 / 4 = 9

9 + 1 = 10

9 - 1 = 8

∴ 36 = 10² - 8²

Here,

16 / 4 = 4

4 + 1 = 5

4 - 1 = 3

∴ 16 = 5² - 3²

So we can take m as 5 and n as 3.

a = 5² - 3² = 25 - 9 = 16

b = 2 x 5 x 3 = 30

c = 5² + 3² = 25 + 9 = 34

∴ 16, 30 and 34 are another Pythagorean triplet.

But we can't consider 16 as c = m² + n² because 16 is not a sum of squares of two counting numbers.

16 = 1 + 15 ( 15 is not a perfect square.)

16 = 4 + 12 (12 is not a perfect square.)

16 = 9 + 7 (7 is not a perfect square.)

16 = 16 + 0 (0 = 0² but 0 can't be taken.)

Hope these may be helpful.

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