1702 A Puzzle Idea from @mathequalslove Tweaked into a Subtraction Puzzle That Directs You to a Post from NebusResearch

Today’s Puzzle:

Joseph Nebus is nearly finished with all the posts in his Little 2021 Mathematics A to Z series. Every year he requests that his readers give him mathematical subjects to write about. At my suggestion, he recently wrote about subtraction, and how it is a subject that isn’t always as elementary as you might expect.  With a touch of humor, we learn that subtraction opens up whole new topics in mathematics.

I wanted to make a puzzle to commemorate his post. I gave it some thought and remembered a tweet from Sarah Carter @mathequalslove:

That puzzle originated from The Little Giant Encylopedia of Puzzles by the Diagram Group. I wondered how the puzzle would work if it were a subtraction puzzle instead of an addition puzzle, and here’s how I tweaked it:

 

There is only one solution. I hope you will try to find it! If you would like a hint, I’ll share one at the end of this post.

Factors of 1702:

  • 1702 is a composite number.
  • Prime factorization: 1702 = 2 × 23 × 37.
  • 1702 has no exponents greater than 1 in its prime factorization, so √1702 cannot be simplified.
  • The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1702 has exactly 8 factors.
  • The factors of 1702 are outlined with their factor pair partners in the graphic below.

More About the Number 1702:

1702 is the hypotenuse of a Pythagorean triple:
552-1610-1702, which is (12-35-37) times 46.

1702² = 2896804, and
2197² = 4826809.
Do you notice what OEIS.org noticed about those two square numbers?

Puzzle Hint:

Here’s how I solved the puzzle: I let the rightmost box be x. Then using the values in the adjacent triangles and working from right to left, I wrote the values of the other boxes in terms of x.

x – 5 went in the box that is second to the right,
x – 5 + 2 = x – 3 went in the next box,
x – 3 + 5 = x + 2,
x + 2 – 6 = x – 4,
x – 4 + 5 = x + 1, and so on until I had assigned a value in terms of x for every box.

Think about it, and this hint should be enough for you to figure out where the numbers from 1 to 9 need to go.

1701 Is a Decagonal Number

Today’s Puzzle:

There is a pattern to the decagonal numbers. Can you figure out what it is?

Factors of 1701:

1701 is divisible by nine because 1 + 7 + 0 + 1 = 9.

  • 1701 is a composite number.
  • Prime factorization: 1701 = 3 × 3 × 3 × 3 × 3 × 7, which can be written 1701 = 3⁵ × 7.
  • 1701 has at least one exponent greater than 1 in its prime factorization so √1701 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1701 = (√81)(√21) = 9√21.
  • The exponents in the prime factorization are 5 and 1. Adding one to each exponent and multiplying we get (5 + 1)(1 + 1) = 6 × 2 = 12. Therefore 1701 has exactly 12 factors.
  • The factors of 1701 are outlined with their factor pair partners in the graphic below.

More About the Number 1701:

1701 is the difference of two squares in SIX different ways.
851² – 850² = 1701,
285² – 282² = 1701,
125² – 118² = 1701,
99² – 90² = 1701,
51² – 30² = 1701, and
45² – 18² = 1701.

1701 is the 21st decagonal number because
21(4·21 – 3) =
21(84-3) =
21(81) = 1701.

There is decagonal number generating function:
x(7x+1)/(1-x)³ = x + 10x² + 27x³ + 52x⁴ + 85x⁵ + . . .

The 21st term of that function is 1701 x²¹.

 

1700 Time for a Horse Race!

Today’s Puzzle:

I’ve made a table of all the numbers from 1601 to 1700, their prime factorizations, and how many factors each of those numbers has.

Each number from 1601 to 1700 has 2, 3, 4, 6, 8, 10, 12, 16, 18, 24, 30 or 40 factors. Do more numbers have 2 factors, 3 factors, or a different number? Sure, you could use the table to count, but it will be more fun if we make a horse race out of it. What pony will you pick, 2, 3, 4, 6? Make your prediction of the winner, and then watch the horse race to see if you were right.

Here are all the horses lined up at the gate. Pick your pony!

And they’re off!

1601 to 1700 Horse Race

make science GIFs like this at MakeaGif
The first half of the race was VERY interesting to me. Three different horses held the lead, and the leaders in the race was even neck and neck some of the time. After you know the winner, watch the race again, following the winner from start to finish.

Factors of the Number 1700:

  • 1700 is a composite number.
  • Prime factorization: 1700 = 2 × 2 × 5 × 5 × 17, which can be written 1700 = 2² × 5² × 17.
  • 1700 has at least one exponent greater than 1 in its prime factorization so √1700 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1700 = (√100)(√17) = 10√17.
  • The exponents in the prime factorization are 2, 2, and 1. Adding one to each exponent and multiplying we get (2 + 1)(2 + 1)(1 + 1) = 3 × 3 × 2 = 18. Therefore 1700 has exactly 18 factors.
  • The factors of 1700 are outlined with their factor pair partners in the graphic below.

More About the Number 1700:

1700 is the sum of two squares THREE different ways:
40² + 10² = 1700,
38² + 16² = 1700, and
32² + 26² = 1700.

Those sum of two squares mean 1700 is the hypotenuse of some Pythagorean triples, SEVEN to be exact:

  1. 260-1680-1700, which is 20 times (13-84-85),
  2. 348-1664-1700, which is 4 times (87-416-425), but it can also be calculated from 32² – 26², 2(32)(26), 32² + 26²
  3. 476-1632-1700, which is (7-24-25) times 68,
  4. 720-1540-1700, which is 20 times (36-77-85),
  5. 800-1500-1700, which is (8-15-17) times 100, but it can also be calculated from 2(40)(10), 40² – 10², 40² + 10²,
  6. 1020-1360-1700, which is (3-4-5) times 340,
  7. 1188-1216-1700, which is 4 times (297-304-425), but it can also be calculated from 38² – 16², 2(38)(16), 38² + 16².

As OEIS.org informs us, 1700 is a Catalan number. It is found in the C(13,4) position, as shown below. With the exception of the lone 1 in the 0th row of the triangle, every number in Catalan’s triangle is the sum of the number above it and the number to its left. For example, 1700 = 1260 + 440.

I hope you have enjoyed learning about the number 1700.

1695 If You Can Solve 3×3 and 5×5 Magic Squares, Then You Can Solve a 15×15 Magic Square!

Today’s Puzzle:

Completing a 15 × 15 magic square may seem daunting, but I assure you, if you can solve a 3 × 3 magic square and a 5 × 5 magic square, then you can complete a 15 × 15 magic square!

The 15 × 15 magic square below is made with twenty-five 3 × 3 magic squares. See the most famous 3 × 3 square in yellow? Do you see that the first 25 multiples of 9 are along the diagonals I’ve drawn? Do you understand the pattern that was used to make this magic square? Study it, then without looking at this one, can you make your own? Open this Excel file, 12 Factors 1683-1695, enable editing, and the sums of each row, column, and diagonal will automatically populate as you type in the numbers.

This next 15 × 15 magic square is made with nine 5 × 5 magic squares. The one in yellow is one of MANY possible 5 × 5 squares. Do you see the first nine multiples of 25 along the diagonals I’ve drawn? Once you understand this pattern, perhaps you would like to take a turn duplicating it. Try it yourself on that same Excel file, 12 Factors 1683-1695.

Factors of 1695:

  • 1695 is a composite number.
  • Prime factorization: 1695 = 3 × 5 × 113.
  • 1695 has no exponents greater than 1 in its prime factorization, so √1695 cannot be simplified.
  • The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1695 has exactly 8 factors.
  • The factors of 1695 are outlined with their factor pair partners in the graphic below.

More About the Number 1695:

Since 1695 is the sum of three consecutive numbers, it is the magic sum of a particular 3 × 3 Magic Square. Those three consecutive numbers are in yellow in the square below.

Also, since 1695 is the sum of the five consecutive numbers shown in yellow below, it is the magic sum of this 5 × 5 magic square:

And lastly, 1695 is the sum of the 15 consecutive numbers shown in yellow below, so here is yet another 15 × 15 magic square with 1695 as the magic sum. This magic square uses the same pattern that works for all odd number magic squares. It is so satisfying to complete it yourself. Study this one and then give it a try! Open that same Excel file, 12 Factors 1683-1695, to make Excel to all the adding for you.

1695 is also the hypotenuse in FOUR Pythagorean triples:
225-1680-1695, which is 15 times (15-112-113),
828-1479-1695, which is 3 times (276-493-565)
1017-1356-1695, which is (3-4-5) times 339, and
1188-1209-1695, which is 3 times (396-403-565).

Did you notice that 565, 339, or 113 was a center number in every magic square in this post?

I hope you enjoy completing these magic squares on your own as you explore the number 1695.

1692 A Pilgrim’s Belt to Unbuckle

Today’s Puzzle:

The logic needed to unbuckle this Pilgrim’s belt puzzle has several interesting twists and turns in it. Even adults will find it a challenge. Guessing and checking will only frustrate you. Use logic to write the numbers from 1 to 10 in both the first column and the top row so that those numbers and the given clues work like a multiplication table.

Here’s the same puzzle if you want to print it using less ink.

Factors of 1692:

  • 1692 is a composite number.
  • Prime factorization: 1692 = 2 × 2 × 3 × 3 × 47, which can be written 1692 = 2² × 3² × 47.
  • 1692 has at least one exponent greater than 1 in its prime factorization so √1692 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1692 = (√36)(√47) = 6√47.
  • The exponents in the prime factorization are 2, 2, and 1. Adding one to each exponent and multiplying we get (2 + 1)(2 + 1)(1 + 1) = 3 × 3 × 2 = 18. Therefore 1692 has exactly 18 factors.
  • The factors of 1692 are outlined with their factor pair partners in the graphic below.
  • 1568 is a composite number.

More About the Number 1692:

1692 is the difference of two squares in three different ways:
424² – 422² = 1692,
144² – 138² = 1692, and
56² – 38² = 1692.

The square of 1692 looks a little interesting:
1692² = 2862864.

Do These Records Refer to One Person, Two People, or More?

Some of my husband’s ancestors lived in Dévaványa, Hungary in the 1700s. I came across a death record for an individual who died in Dévaványa but was born in Gyoma, the place where two of my husband’s grandparents were born.

Dévaványa civil death record, number 323, 1899 October 24, Csáki Gergely, age 73 (born about 1827) in Gyoma, married to Karádi Eszter, his parents were the late Csáki István and the late Kis Rebeka.

My husband had a 4th great-grandmother named Kis Erzsébet who lived in Gyoma. Perhaps this Csáki Gergely would be related to her somehow? I decided to look for him in other records and found that he and his wife had one child born in Gyoma.

Gyoma Reformed Church christening record, page 877, Line 188, born 1859 August 6, christened 1859 August 8, a girl, Ester, born to Csáki Gergely and Karádi Ester.

That matches the information in Csáki Gergely’s death record. Great! Let me see if I can find his christening record. I’m looking for a Gergely who was born about 1827 and his parents were Csáki István and Kis Rebeka. I did find a Gergely christened on 12 September 1827 whose father was Csáki István, but his mother’s name was Fekete Rebeka. Also that Gergely died a half year later on 25 April 1828. The couple had another son, István, christened 23 Jan 1829, and one more they named Gergely christened on 12 January 1830. Could that second Gergely be the same as the one who died in 1899?

Suppose one of your relatives died and you reported the death to the authorities. Would you get his/her birth year and mother’s maiden name right? I’m not ready to declare that these two documents refer to the same person, but I’m also not going to eliminate the possibility either.

Some people in Hungary essentially had two surnames. Perhaps Rebeka sometimes listed her surname as Kis. I searched for István and Rebeka’s marriage record.

Gyoma Reformed Church marriage record, page 1009, last entry for 1826 November 15, the widower young Csáki István, age 25 (born about 1801) and Fekete György’s daughter Rebeka, age 29 (born about 1797).

There is no mention of Kis there, and young Csáki István was already a widower? I looked for Rebeka’s christening record and found she appears to have been christened on 13 Jun 1797, the daughter of Vanyai Fekete György and Szabó Ilona. (Vanyai means he was from Dévaványa.) Still no mention of the surname Kis. I found these parents’ marriage record.  Gyoma Reformed Church marriage record, 2nd entry for 29 January 1794, Dévaványa resident Fekete György, age 25 (born about 1769), and the late Bálint Szabó Márton’s daughter Ilona, age 21 (born about 1773). Again, no mention of Kis on either her father’s side or her mother’s side. It looks like I’m at a dead end.

All this means Csáki Gergely might not be related to my husband’s 4th great-grandmother after all, but my curiosity was still peaked about Csáki István being a widower at age 25. I found three records of interest.

  1. Gyoma Reformed Church marriage record page 1000, 16 November 1825, Csáki Gergely’s son István, age 24, wed B. Szabó János’s daughter, Mária, age 17 (born about 1808). This is one day less than one year before the widower Csáki István married Fekete Rebeka. And her name is B. Szabó Mária? Could she be related to Fekete Rebeka’s mother, Bálint Szabó Ilona?
  2. Gyoma Reformed Church christening record, page 73, 21 September 1826, the child christened, Imre, his parents were Csáki István and Kaptsos Mária.  Mária is the right given name, but Kaptsos and B. Szabó are not the same names.
  3. Gyoma Reformed Church death record, page78, 22 September 1826, Csáki Istvánné (Mrs. István Csáki), age 24 (born about 1802). This is just one day after the christening mentioned in the previous record. I looked up the cause of death, pokolvar, to see if it had anything to do with childbirth, but it did not. Pokol means hell, pokolvar means carbuncles. I do not know if the two words are related. Her death occurred less than two months before Fekete Rebeka’s 15 November 1826 wedding mentioned above. (It wasn’t unusual for a father of a new baby to look for a new mother soon after the death of his wife.)

Do these three records belong to one person, two people, or three? I looked for a possible christening record for Mária, and found several records for children born to János (Kaptsos or Bálint) and Ersebet Czegledi. They may be one couple or they may be two, but I arranged the information from the records in a chart here:

Note that Győri István was the godfather of three of the children. Balás Judit (not Ersébet) might have been the godmother all three times as well.

Discrepancies might have been from the participants, the informants, or even the priests. Most people were illiterate back then. I can imagine a person going to the priest when Mrs. István Csáki died. “How old was she?” asks the priest. “I don’t know,” the informant responds. “What were her parents’ names? I’ll determine her age from the christening book,” says the priest. He finds the birth of Mária in 1802, totally unaware that Mária 1802 died in 1807. “It looks like she was 24 when she died,” and that’s what he records in the book. Another possibility is that all the records are exactly correct and belong to different people so there are no discrepancies.

Although I was very interested in looking at these records, and I have an opinion of which records refer to which people, I’m not offering any of those opinions. But if you are related to any of these people, and find my research helpful, go ahead, form your own opinions, and make any connections you feel are appropriate.

 

1691 Jack 0’Lantern Time

Today’s Puzzle:

This smiling Jack O’Lantern actually has a few tricks up his sleeve. Don’t let yourself get fooled by any of them. Using logic, write the numbers from 1 to 12 in the first column as well as in the top row so that those numbers and the given clues can be a multiplication table.

Have a safe and happy Halloween!

Here’s the same puzzle with just the clues:

Factors of 1691:

  • 1691 is a composite number.
  • Prime factorization: 1691 = 19 × 89.
  • 1691 has no exponents greater than 1 in its prime factorization, so √1691 cannot be simplified.
  • The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1691 has exactly 4 factors.
  • The factors of 1691 are outlined with their factor pair partners in the graphic below.

More About the Number 1691:

1691 is the hypotenuse of a Pythagorean triple:
741-1520-1691, which is 19 times (39-80-89).

1684 Triangular Candy Corn

Today’s Puzzle:

Candy corn is a triangular piece of Halloween candy. 1684 is a centered triangular number formed from the sum of the 32nd, the 33rd, and the 34th triangular numbers. Label the boxes next to the representations of each of those triangular numbers.

 

Factors of 1684:

  • 1684 is a composite number.
  • Prime factorization: 1684 = 2 × 2 × 421, which can be written 1684 = 2² × 421.
  • 1684 has at least one exponent greater than 1 in its prime factorization so √1684 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1684 = (√4)(√421) = 2√421.
  • The exponents in the prime factorization are 2 and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1) = 3 × 2 = 6. Therefore 1684 has exactly 6 factors.
  • The factors of 1684 are outlined with their factor pair partners in the graphic below.

More About the Number 1684:

1684 is the sum of two squares:
30² + 28² = 1684.

1684 is the hypotenuse of a Pythagorean triple:
116-1680-1684, calculated from 30² – 28², 2(30)(28), 30² + 28².
It is also 4 times (29-420-421).

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

1680 square roots

1684/2 = 842,  which is the third number in the smallest set of FIVE consecutive numbers whose square roots can be simplified.

1682 This Puzzle Is Not as Difficult as It Looks

Today’s Puzzle:

Three months ago I was inspired by a puzzle I saw on Twitter:

I enjoyed solving this complicated-looking system of equations, but let me tell you, Looks Can Be Deceiving! The puzzle is not as difficult as it looks.

I decided to make a similar puzzle, and I’ve waited for my 1682nd post to share it with you. If you can solve the Twitter puzzle, then you can solve my puzzle, too!

Why did I wait until my 1682nd post to share this puzzle? Because if you add the three equations together you get:
(x + y + y + z + x + z)(x + y + z) = 1682,
(2x + 2y + 2z)(x + y + z) = 1682,
2(x + y + z)(x + y + z) = 1682,
2(x + y + z)² = 1682.
The factors of 1682 will be quite helpful at this point. What is the greatest common factor of the numbers after the equal signs?

The numbers in one of 1682’s Pythagorean triples, 580-609-1682, are featured prominently in this puzzle.

I hope you enjoy solving my puzzle, and maybe you will make and solve some puzzles of your own!

Factors of 1682:

  • 1682 is a composite number.
  • Prime factorization: 1682 = 2 × 29 × 29, which can be written 1682 = 2 × 29².
  • 1682 has at least one exponent greater than 1 in its prime factorization so √1682 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1682 = (√841)(√2) = 29√2.
  • The exponents in the prime factorization are 1 and 2. Adding one to each exponent and multiplying we get (1 + 1)(2 + 1) = 2 × 3 = 6. Therefore 1682 has exactly 6 factors.
  • The factors of 1682 are outlined with their factor pair partners in the graphic below.

More About the Number 1682:

1682 is the sum of two squares in two different ways:
29² + 29² = 1682, and
41² + 1² = 1682.

1682 is the hypotenuse of two Pythagorean triples:
82-1680-1682, calculated from 2(41)(1), 41² – 1², 41² + 1², and
1160-1218-1682, which is (20-21-29) times 58.

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

1680 square roots

1682/2 = 841, which is the second number in the smallest set of FIVE consecutive numbers whose square roots can be simplified.

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!

Duncan Fraser

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.

ivasallay

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!

Duncan Fraser

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.

1680 square roots

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.