### Can You See How 779’s Factor Pairs Are Hiding in Some Pythagorean Triples?

- 779 is a composite number.
- Prime factorization: 779 = 19 x 41
- The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 779 has exactly 4 factors.
- Factors of 779: 1, 19, 41, 779
- Factor pairs: 779 = 1 x 779 or 19 x 41
- 779 has no square factors that allow its square root to be simplified. √779 ≈ 27.91057.

Those factor pairs are hiding in some Pythagorean triples. Scroll down to read how, but first here’s today’s puzzle:

Print the puzzles or type the solution on this excel file: 10-factors-2016

————————————–

And now some Pythagorean triple number theory using 779 as an example:

The factors of 779 are very well hidden in five Pythagorean triples that contain the number 779**.** Here’s how: 779 has two factor pairs: **19** x **41 **and **1** x **779**. Those factor pairs show up in some way in each of the calculations for these 779 containing Pythagorean triples:

- 171-760-
**779**which is**19**times each number in 9-40-**41**. **779**-303420-303421, a primitive calculated from**779**(**1**); (**779**² –**1**²)/2; (**779**² +**1**²)/2.**779**-7380-7421 which is**41**times each number in**19**-180-181.**779**-15960-15979 which is**19**times each number in**41**-840-841.- 660-
**779**-1021, a primitive calculated from (**41**² –**19**²)/2;**19**(**41**); (**41**² +**19**²)/2.

Being able to find whole numbers that satisfy the equation a² + b² = c² is one reason why finding factors of a number is so worth it. ANY factor pair for numbers greater than 2 will produce at least one Pythagorean triple that satisfies a² + b² = c². The more factor pairs a number has, the more Pythagorean triples will exist that contain that number. 779 has only two factor pairs so there are a modest number of 779 containing Pythagorean triples. All of its factors are odd so it was quite easy to find all of the triples. Here’s a brief explanation on how each triple was found:

- 799 has one prime factor that has a remainder of 1 when divided by 4. That prime factor, 41, is therefore the hypotenuse of a primitive Pythagorean triple. When the Pythagorean triple is multiplied by the other half of 41’s factor pair, 19, we get a Pythagorean triple in which 779 is the hypotenuse.
- Every odd number greater than 1 is the short leg of a primitive Pythagorean triple. To find that primitive for a different odd number, simply substitute the desired odd number in the calculation in place of 779.
- Because every odd number greater than 1 is the short leg of a primitive Pythagorean triple,
**1****9**(**1**); (**1****9**² –**1**²)/2; (**19**² +**1**²)/2 generates the primitive triple (**19**-180-181). Multiplying each number in that triple by the other half of**19**‘s factor pair,**41**, produces a triple with**779**as the short leg. - Because every odd number greater than 1 is the short leg of a primitive Pythagorean triple,
**41**(**1**); (**41**² –**1**²)/2; (**41**² +**1**²)/2 generates the primitive triple (**41**-840-841). Multiplying each number in that triple by the other half of**41**‘s factor pair,**19**, produces a triple with**779**as the short leg. - Since factor pair
**19**and**41**have no common prime factors, the formula (**41**² –**19**²)/2;**19**(**41**); (**41**² +**19**²)/2 produces another primitive triple 660-**779**-1021. If they did have common factors, the factor pair would still produce a triple, but it would not be a primitive one.

Here’s some other interesting facts about the number 779:

779 is the sum of eleven consecutive prime numbers:

47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97 = 779.

779 can also be written as the sum of three squares six different ways:

- 27² + 7² + 1² = 779
- 27² +
**5**² +**5**² = 779 - 23² + 15² + 5² = 779
- 23² + 13² + 9² = 779
- 21² + 17² + 7² = 779
- 21² +
**13**² +**13**² = 779

Finally, the table below shows some logical steps that could be used to solve Puzzle #779:

## Recent Comments