We all know that Pythagorian Triplet must be a whole number and must be \(\sqrt{a^2 + b^2} = c \). Below is a list of some Pythagorian Triplet ( \(a\) is the smallest leg and \(c\) is the largest leg or the hypotenuse)...

\(a\) | \(b\) | \(c\) |

3 | 4 | 5 |

9 | 40 | 41 |

6 | 8 | 10 |

15 | 36 | 39 |

But did you know that you can predict Pythagorian Triplet using only the smallest leg? But before that, we must identify what is SIMPLE TRIPLET and PRODUCT TRIPLET. What are those? Simple triplet is a Pythagorean Triplet whose values are not multiplied with any number while Product Triplet is a Pythagorean Triplet whose values are products of values of a simple triplet and a number.

From the above example, the Simple Triplets there are 3, 4, 5 and 9, 40, and 41 while the Product Triplet are 6, 8, 10 and 15, 36, 39. Why? You cannot lower the terms if Simple Triplet (like 3, 4, 5) is in ratio (3:4:5) while in Product Triplet, you can. You can lower 6:8:10 into 3:4:5 and 15:36:39 into 5:12:13.

Another way of identifying SimpleTriplet and Product Triplet is subtracting the larger leg from the hypotenuse (\(c - b\)). If \(c - b = 1\), then it is a Simple Triplet. If \(c - b > 1\), then it is a Product Triplet. (Note: the value of \(c - b\) of a Product Triplet is the number to be divided from the triplet to get the Simple Triplet)

Example: From first triplet (3, 4, 5) \[c - b = ?\] \[5 - 4 = 1\] \(c - b = 1\), then it is a Simple Triplet

From fourth example (15, 36, 39) \[c - b = ?\] \[39 - 36 = 3\] \(c - b > 1\), then it is a Product Triplet

In which Simple Triplet it was based? (Note: \(a_p\) is \(a\) of Product Triplet while \(a_s\) is \(a\) of the Simple Triplet. Same as \(b\) and \(c\))

For the value of \(a_s\) \[a_p = (a_s)(c - b)\] \[15 = (a_s)(3)\] \[\frac{15}{3} = \frac {(a_s)(3)}{3}\] \[a_s = 5\]

For the value of \(b_s\) \[b_p = (b_s)(c - b)\] \[36 = (b_s)(3)\] \[\frac{36}{3} = \frac {(b_s)(3)}{3}\] \[b_s = 12\]

For the value of \(c_s\) \[c_p = (c_s)(c - b)\] \[39 = (c_s)(3)\] \[\frac{39}{3} = \frac {(c_s)(3)}{3}\] \[c_s = 13\]

It was based from the Simple Triplet \(5, 12,\) and \(13\)

Now, how could you identify using only the smallest leg (\(a\))given? By using the formula \(b = \frac {a^2 - 1}{2}\) and \(c = \frac {a^2 - 1}{2} + 1\) or simply \(c = b + 1\), you can now get the Simple Triplet.

For example: If \(a = 3\), what is \(b\) and \(c\)?

Using the formula for \(b\), \[b = \frac {a^2 - 1}{2}\] \[b = \frac {(3)^2 - 1}{2}\] \[b = \frac {9 - 1}{2}\] \[b = \frac {8}{2}\] \[b = 4\]

and formula for \(c\) \[c = b + 1\] \[c = 4 + 1\] \[c = 5\]

The values for \(b\) and \(c\) are \(4\) and \(5\), respectively.

Other example:

- If \(a = 7\), what is \(b\) and \(c\)?

Using the formula for \(b\), \[b = \frac {a^2 - 1}{2}\] \[b = \frac {(7)^2 - 1}{2}\] \[b = \frac {49 - 1}{2}\] \[b = \frac {48}{2}\] \[b = 24\]

and formula for \(c\) \[c = b + 1\] \[c = 24 + 1\] \[c = 25\]

Let's check using Pythagorian Theorem: \[c = \sqrt {a^2 + b^2}\] \[25 = \sqrt {(7)^2 + (24)^2}\] \[25 = \sqrt {49 + 576}\] \[25 = \sqrt {625}\] \[\boxed{25 = 25} \]

Limitations for these formulas:

It can be applicable if the value of \(a\) is odd.

It can be applicable if the value of \(a\) is a whole number.

(Honestly, I just discovered these formulas without reading. I hope it works)

## Comments

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TopNewestConsider \(a = 65.\) By your formulae, \[b = 2112,\ c = 2113.\] If you check the numbers, it turns out that yes, they work. So the formulae are useful, at least for coming up with triples. But what if I told you there is a significantly smaller simple triple for \(a = 65?\) \[b = 72,\ c = 97.\] So you can't actually predict a pythagorean triple given only the smallest leg, you can just find one simple solution in particular. – Caleb Townsend · 2 years ago

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– Jade Mijares · 2 years ago

You can predict, but not all triples...Log in to reply

– Caleb Townsend · 2 years ago

That is precisely the point; you can find a pythagorean triple with those formulae, but if I told you to guess the triple I'm thinking of given the smallest side length, you won't necessarily guess my triple.Log in to reply

– Jade Mijares · 2 years ago

I'm not a mind reader to guess. Many formulas have their own limitations. I know only few is applicable in these formulas. If you tend to prove that I am wrong, then I can say that the triples you are showing are out of my topic.Log in to reply

– Caleb Townsend · 2 years ago

To clarify, there are not only additional product triplets that can be found, but often additional simple triplets as well, that do not obey the formulae.Log in to reply

– Jade Mijares · 1 year, 4 months ago

I understand now.... Thank you!!!Log in to reply