The most famous Pythagorean triple is 3-4-5.
Perhaps you know that of the three numbers in EVERY Pythagorean triple at least one of them will be divisible by 3, at least one of them will be divisible by 4, and at least one of them will be divisible by 5. That’s obvious for triple 3-4-5.
Here’s another example: Pythagorean triple 11-60-61. Two of those numbers are prime numbers, yet 60 is divisible by 3, 4, AND 5.
If the Pythagorean triple is a primitive, something else amazing happens:
That’s amazing all by itself, but that’s only part of the picture. Let’s look at the complete picture specifically using the triple with hypotenuse 821:
- 25² + 14² = 821
- 429² + 700² = 821²
- 429–700–821 can be calculated from 25² – 14², 2(25)(14), 25² + 14²
- 821 + 700 = 1521
- √1521 = 39 which is 25 + 14
- 429 ÷ 39 = 11
WHY does this happen to primitive Pythagorean triples?
It happened for 429–700–821 because 700 + 821 = 25² + 14² + 2(25)(14) = (25 + 14)².
After all (a + b)² = a² +2ab + b² is always true.
And it can be rearranged: (a + b)² = a² + b² +2ab.
You can use a similar proof whenever the element of the Primitive triple that is divisible by 4 can be expressed as 2ab.
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But it is a different, and perhaps simpler, story for many triples such as 9–40–41 which was calculated using (9)(1), (9² – 1²)/2, (9² + 1²)/2.
In that case 40 + 41 = (9² – 1²)/2 + (9² + 1²)/2 = (9² – 1² + 9² + 1²)/2 = 9².
You can use a similar proof whenever the element of the Primitive triple that is divisible by 4 can be expressed as (a² – b²)/2.
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Go ahead, take ANY primitive Pythagorean triple. Add the leg that is divisible by 4 to its hypotenuse. You will get a perfect square. Here’s a few more examples:
Similarly primitive triples that have been multiplied by a square number will also produce a perfect square, but you’ll have to be careful which leg you add to the hypotenuse if you multiplied by a square number that is a multiple of 4.
For example, 9(3-4-5) = 27-36-45 and 36 + 45 = 81, a square number.
But 4(3-4-5) = 12-16-20 so each number is divisible by 4. Note that 16 + 20 = 36, a square number, but 12 + 20 does not.
Here’s a few other essential facts about the number 821:
All of the odd numbers between 820 and 830, except 825, are prime numbers. Thus 821, 823, 827, 829 is the fourth prime decade.
- 821 is a prime number.
- Prime factorization: 821 is prime.
- The exponent of prime number 821 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 821 has exactly 2 factors.
- Factors of 821: 1, 821
- Factor pairs: 821 = 1 x 821
- 821 has no square factors that allow its square root to be simplified. √821 ≈ 28.65309756.
How do we know that 821 is a prime number? If 821 were not a prime number, then it would be divisible by at least one prime number less than or equal to √821 ≈ 28.7. Since 821 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 821 is a prime number.
Here’s another way we know that 821 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 25² + 14² = 821 with 25 and 14 having no common prime factors, 821 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √821 ≈ 28.7. Since 821 is not divisible by 5, 13, or 17, we know that 821 is a prime number.
No, I hadn’t noticed that !*! Thanks for another intriguing mathsbit/mathbit worth proving.
You’re very welcome. I’ve learned things about Pythagorean triples from your website, too.
The sum of the even leg and hypotenuse for all ppts (primitive Pythagorean triples) is the square of an odd number. This is a rule for ppts that has been sadly overlooked. Because of this, you can find several ppts that have the sum of the even leg and hypotenuse with the same numerical value. Example (35,12,37), (21,20,29), (7,24,25). If their sum is the square of an odd number, p, the number of ppts is (p-1)/2 ppts. If the sum of the square of an odd number is not prime, then (p-1)/2 Pythagorean triples are produced.
Some are ppts and the others are scalar multiples. Example 15 will produce 4 ppt and 3 scalar multiples.
Since for ppts, the sum of the even leg and the hypotenuse is an odd number squared, and an odd number is the sum of an even and an odd number. (m+n)squared gives the m, n parts for the hypotenuse and even leg of ppts.
Wow, thanks! So since 15² = 225, the seven Pythagorean triples whose even side added to the hypotenuse equals 225 are (197-28-197), (165-52-173), (135-72-153), (105-88-137), (75-100-125), (45-108-117), and (15-112-113). I just had to find them myself after finding this out!
Glad you were able to get the 7 triples. The rule for odd numbers is an odd number is the sum of an even and an odd. I did show that the even side 2mn and hypotenuse m^2+ n^2 is m^2+n^2+2mn which is (m+n)^2, m+n is.odd.
M+N =15 is an equation with two variables. However, each has to be an integer. Thus (m,n) are (14,1) (13,2) (12,3) (11,4) (10,5) (9,6) (8,7). Note 3 sets of(m,n) have gcfs and these will be the non ppts.
The last pair of (8,7) has m=n+1. In your output (15, 112,113) you will note that 15 is now your odd leg., and the difference between the hypotenuse and the even leg is also 1.
This leads to a very interesting rule , which is
If the difference between hypotenuse and even leg of a ppt is1. Then the area divided by the the perimeter of the ppt is (odd leg-1)/4 . Since in your odd leg is 15, the ratio is(15-1)/4= 3.5
Now if the odd leg is a prime number then only one ppt is formed eg( 7,24,25).
All composite or multiples will produce more than one Pythagorean triple ,one of which will produce a ppt with the hypotenuse minus the even leg equal 1,
For example 9 will be ( 9, 40,41) with m= n+1 .The other (9,12,15).
We can simply square 9 so (81+1)/2 =41 and(81-1)/2=40 are the hypotenuse and even leg respectively.
Or To get the the m and n ,m =(x+1)/2, n=(x-1)/2 . Thus x= 9 gives m=5 and n=4
So for the odd legs starting with the series of odd numbers 3,5,7,9…….. and
Applying the above rule where the difference between the hypotenuse and even leg is1
The area/ perimeter = (x-1)/4 we will get an arithmetic series with first term 0.5 and a common difference of 0.5
Using the formula, 3 gives 0.5, 5(1),7(1,5), 9 (2) ………
Thus producing the arithmetic series.
To develop the formula, Let x be the length of the odd leg. The hypotenuse plus the even leg is x^2 with the hypotenuse (x^2-1)/2 and the even leg (x^2-1)/2. Thus the perimeter is x +x^2 or x(x+1)
The area is x (x^2-1)/4 or x(x+1)(x-1)/4
Area divided by perimeter (x-1)/4. This formula can be used by kids as young as grade 7.
The next formula is area/perimeter = n/2. Where is equal to m minus one
You can develop this formula starting with the hypotenuse m^2+n^2 minus the even leg 2mn which leads to (m-n)^2=1 thus (m-n)=1
Since the odd leg is m^2-n^2 which is (m+n)(m-n) =(m+n)
I will leave the rest to any interested person.
Thank you for explaining more. This will be my topic when I write my 1681st post in a couple of months. I will share your comments in that post so hopefully more people will see them. Thanks again!
If the odd leg of a ppt is a prime number where its rightmost digit is 1 or 9 then the even leg is a multiple of 60. As you showed (11, 60, 61). (19, 180, 181). Some of the hypotenuses will also be a prime number. Of course the even leg is a multiple of 4 (2mn) and either m or n is an even number.
The uniqueness of the A^2+B^2 = C^2 formula is that it holds for any A and B and C. The Pythagorean triples are a special case where all three are integers. The amazing truth is for all As, Bs and Cs the three sides of a right triangle where A is greater than B. (a+b)^2 +(a-b)^2 = 2(c)^2
Example 3^2+4^2=5^2 , (3+4)^2 + (4-3)^2= 2(5)^2
Algebraically (a+b)^2+(a-b)^2 is 2a^2+2b^2 which is 2c^2 and thus diving by 2 gives the original formula.
Can exponents greater than 2 produce a similar result.?
Let us assume a^3+ b^3= c^3 a>b. And a a b are integers
Then (a+b)^3+(a-b)^3 is 2a^3 + 6ab^2. Is 6ab^2 = 2b^3?
Thus 6ab^2-2b^3=0 and 2b^2(3a-b)=0. Either b=0 or b = 3a. If b=0 then a^3=c^3. If b=3a then a^3+(3a)^3=28a^3=c^3
The cube root of 28 is not and integer and thus c is not and integer.
Let us consider a^4+b^4=c^4 then
(a+b)^4 +(a-b)^4 = 2a^4+12a^2b^2+2b^4. The extra term must be equal to zero to have the equation equal to 2c^4. Thus either a =0 or b=0. Thus the extra term shows that c cannot be an integer in the original fourth degree equation.
All higher degrees behave like n equal 3 for odd exponents and n equal 4 for even exponents.
Did Fermat have a simple proof. He did not have the tools available to Prof. Wiles