# What Is 1681’s Claim to Fame?

### Today’s Puzzle:

1681 shares that claim to fame with those other numbers, but is the only one on the list that ______________________________________________________________.

### Why Do Pythagorean Triples Do What They Do?

Four years ago I noticed something remarkable about Pythagorean triples and wrote a post that I titled Why Do Pythagorean Triples Do That?

This graphic shows some of what I found so amazing:

It happens over and over again because (a + b)² = a² +2ab + b² is always true.

That statement can be rearranged: (a + b)² = a² + b² +2ab.

Here are some more examples:

Earlier THIS year, Duncan Fraser found my post and left a series of comments that are just too good to keep to myself. I decided I would share his comments the next time I wrote about an odd perfect prime, which 1681 happens to be. Here is our conversation (I’ve replaced ^ with exponents to make reading it a little easier.):

Duncan Fraser

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 m²+ n² is m²+n²+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 the hypotenuse and even leg of a ppt is1. Then the area divided by the perimeter of the ppt is (odd leg-1)/4. Since 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 x² with the hypotenuse (x²-1)/2 and the even leg (x²-1)/2. Thus the perimeter is x +x² or x(x+1)
The area is x (x²-1)/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 m²+n² minus the even leg 2mn which leads to (m-n)²=1 thus (m-n)=1

Since the odd leg is m²-n² 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 should have said 7 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 A²+B² = C² 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 3²+4²=5² , (3+4)² + (4-3)²= 2(5)²

Algebraically (a+b)²+(a-b)² is 2a²+2b² which is 2c² and thus diving by 2 gives the original formula.

Can exponents greater than 2 produce a similar result.?

Let us assume a³+ b³= c³ a>b. And a and b are integers

Then (a+b)³+(a-b)³ is 2a³ + 6ab². Is 6ab² = 2b³?

Thus 6ab²-2b³=0 and 2b²(3a-b)=0. Either b=0 or b = 3a. If b=0 then a³=c³. If b=3a then a³+(3a)³=28a³=c³
The cube root of 28 is not an integer and thus c is not an integer.

Let us consider a⁴+b⁴=c⁴ then

(a+b)⁴ +(a-b)⁴ = 2a⁴+12a²b²+2b⁴. The extra term must be equal to zero to have the equation equal to 2c⁴. 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.

That was wonderfully proven!

### A Puzzle with Pythagorean Triples:

Whether or not you were able to follow all that Duncan Fraser wrote, fill in the blanks on this next puzzle. Seriously, you should be able to complete it in less than a couple of minutes!

Now sit back, relax, take notice and wonder about the patterns in this table:

Mathematics is truly a thing of beauty!

### Factors of 1681:

• 1681 is a composite number and is a perfect square.
• Prime factorization: 1681 = 41 × 41 which can be written 1681 = 41².
• 1681 has at least one exponent greater than 1 in its prime factorization so √1681 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1681 = (√41)(√41) =
• The exponent in the prime factorization is 2. Adding one to that exponent, we get (2 + 1) = 3. Therefore 1681 has exactly 3 factors.
• The factors of 1681 are outlined with their factor pair partners in the graphic below.

### More About the Number 1681:

1681 is the sum of two squares:
40² + 9² = 1681.

1681 is the hypotenuse of two Pythagorean triples:
369-1640-1681, which is 41 times (9-40-41).
720-1519-1681, calculated from 2(40)(9), 40² – 9², 40² + 9².

1680, 1681, 1682, 1683, and 1684 are the second smallest set of FIVE consecutive numbers whose square roots can be simplified.

### Puzzle Solution:

All of the numbers are perfect squares AND concatenations of exactly two perfect squares:
7² = 49; 2² = 4, 3² = 9.
13² = 169; 4² = 16, 3² = 9.
19² = 361; 6² = 36, 1² = 1.
35² = 1225; 1² = 1, 15² = 225.
38² = 1444; 12² = 144, 2² = 4.
41² = 1681; 4² = 16, 9² = 81.
57² = 3249; 18² = 324, 3² = 9.
65² = 4225; 2² = 4, 15² = 225.
70² = 4900; 2² = 4, 30² = 900.

1681 shares that claim to fame with those other numbers, but is the only one on the list that is a concatenation of two 2-digit squares.

# 269 and Five More Consecutive Square Roots

• 269 is a prime number.
• Prime factorization: 269 is prime.
• The exponent of prime number 269 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 269 has exactly 2 factors.
• Factors of 269: 1, 269
• Factor pairs: 269 = 1 x 269
• 269 has no square factors that allow its square root to be simplified. √269 ≈ 16.401

How do we know that 269 is a prime number? If 269 were not a prime number, then it would be divisible by at least one prime number less than or equal to √269 ≈ 16.401. Since 269 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 269 is a prime number.

As I have previously written, 844, 845, 846, 847, and 848 are the smallest FIVE consecutive numbers whose square roots can be simplified. Here are the second smallest FIVE with the same property.

The first number in the second set, 1680, equals 2 x 840 which is very close to the first number in the first set. Will strings of five consecutive numbers with reducible square roots occur about once every 850 numbers?

We can find the number of factors for these numbers by examining their prime factorizations.

The number of factors for each of the integers in this second set ranges from 3 to 40. Only two of the integers have the same number of factors. Finding another string of four or more numbers that have reducible square roots as well as the same number of factors may be difficult.