A Pythagorean triplet is a set of 3 positive integer numbers which may be the sides of a right triangle, i.e. they meet the Pythagorean theorem c² = a² + b².
You can check that the numbers on your table are Pythagorean triplets by substituting them in the Pythagorean equation:
- 6² + 8² = 36 + 64 = 100 = 10²
- 8² + 15² = 64 + 225 = 289 = 17²
- 10² + 24² = 100 + 576 = 676 = 26²
- 12² + 35² = 144 + 1225 = 1369 = 37²
Now, lets look for the pattern:
x-value Pythagorean
triple
3 (6, 8, 10) 6/2 = 3
3² - 1 = 9 - 1 = 8
3² + 1 = 9 + 1 = 10
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4 (8, 15, 17) 8/2 = 4
4² - 1 = 16 - 1 = 15
4² + 1 = 16 + 1 = 17
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5 (10, 24, 26) 10/2 = 5
5² - 1 = 25 - 1 = 24
24² + 1 = 25 + 1 = 26
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6 (12, 35, 37) 12/2 = 6
6² - 1 = 36 - 1 = 35
6² + 1 = 36 + 1 = 37
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From which you find the pattern: the first number is 2x, the second number is x² - 1, and the third number is x² + 1
⇒ (2x)² + (x² - 1)² = (x² + 1)², or
(x² - 1)² + (2x)² = (x² + 1)².
Other example of a Pythagorean triple is (3, 4, 5). You migth think that it does not follow the pattern, but if you do x = 2, you end with:
- x = 2
- 2x = 2(2) = 4
- x² - 1 = 2² - 1 = 3
- x² + 1 = 2² + 1 = 5
Hence, (3, 4, 5) also follows the pattern.
Only right triangles with non-integer sides do not form Pythagorean triples.
Of course you may proof that (x² - 1)² + (2x)² = (x² + 1)² is an identity (always true):
Left hand side: (x⁴ - 2x² + 1) + 4x² = x⁴ + 2x² + 1
Right hand side: x⁴ + 2x² + 1
∴ The equation is always true.
At the end, the pattern is true for any Pythagorean triplet, but a more formal proof is beyond the scope of this question.
