Pythagorean Triples

Thus combining all the above results, the nth Pythagorean triple can be written in a more compact form as:

in which n is the nth Pythagorean triple. It is worth mentioning that the above Pythagorean triple is obviously based on certain assumptions:

• the shortest side a and the longest side (hypotenuse) c are always odd numbers while the middle side b is always an even number

• all the sides of any Pythagorean triple are co-prime, i.e. these numbers do not have a common factor.

Proving of formulas

In this I am going to show that my formulas are correct to calculate any Pythagorean triple of which a is an odd number. Let me consider the 40th triple. Therefore n = 40. Using my formulas I calculate the sides of this right-angled triangle:

a = 2n + 1 = 2 × 40 + 1 = 81

b = 2n2 + 2n = 2 × 402 + 2 × 40  = 3280

c = 2n2 + 2n + 1 = 2 × 402 + 2 × 40 + 1  = 3281

I use the above values of (a, b, c) as the 40th Pythagorean triple:

a2 + b2  = 812 + 32802 = 6561 + 10758400 = 10764961

c2 = 32812 = 10764961

Thus it is proved that my formulas work very well to calculate any Pythagorean triple.

I can say from my knowledge on Pythagorean theorem that (6, 8, 10) is also a set of Pythagorean triple because 62 + 82 = 102. But these numbers have a common factor of 2 or they are all multiples of 2. So triples of this type will follow certain pattern of their own, and hence I decided to investigate them separately.

Now that I have all my general formulas for finding each of the triples, it is reasonable to go back to the original theorem, a2 + b2  = c2. As a proof, I can show that the previous formulas satisfy this equation.

(2n + 1)2 + (2n2 + 2n)2 = (2n2 + 2n +1)2

By considering the left-hand side (LHS) and right-hand side (RHS) of the above equation separately, and expanding and simplifying, I get:

LHS:

   (2n + 1)2 + (2n2 + 2n)2 

= 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 

= 4n4 + 8n3 + 8n2 + 4n + 1

RHS:

   (2n2 + 2n + 1)2 

= (2n2 + 2n + 1) (2n2 + 2n + 1)

= 4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n + 2n2 + 2n + 1

= 4n4 + 8n3 + 8n2 + 4n + 1

This clearly shows that LHS = RHS or that the sum of the squares on the shortest and middle sides is equal to the square on the hypotenuse or the longest side. This also shows that the formulas I derived for the sides of any right angle triangle are valid for any Pythagorean triple of which the shortest side is always an odd number.

Finding a formula for the Perimeter P

In this I am going to derive the formula for the perimeter of a right-angled triangle algebraically. The perimeter P of a triangle is found by adding all its sides.

P = a + b + c

Thus the perimeter of the right-angled triangle with the Pythagorean triple of (2n + 1, 2n2 + 2n, 2n2 + 2n + 1) is obtained by substituting for a, b and c from this set of triples as:

P = 2n + 1 + 2n2 + 2n + 2n2 + 2n + 1

   = 4n2 + 6n + 2

   = 2(2n2 + 3n + 1)

Factorising further, I get the formula for the perimeter of the right-angled triangle as:

P = 2(n + 1)(2n + 1)

The above formula can also be derived using the values of the perimeter given in the fifth column of Table 1.

From the above I see that the second difference is constant and hence I will use the equations I have derived in Appendix B to find the nth term of the quadratic sequence. Thus I find that:

2a = 8                (considering the 2nd difference)

  a = 4

3a + b = 18                (considering the 1st term of the 1st difference)

Substituting for a in the above equation        ⇒ 12 + b = 18        ⇒ b = 6

Considering the 1st term of the sequence S1 and substituting for a and b,

a + b + c = 12        ⇒ 4 + 6 + c = 12                ⇒ c = 2

So the formula to find the perimeter P of the right-angled triangle with the Pythagorean triples of (2n + 1, 2n2 + 2n, 2n2 + 2n + 1) is:

P = 4n2 + 6n + 2

    = 2(2n2 + 3n + 1)

Factorising further, I get exactly the same formula as I derived earlier using algebra for the perimeter as:

P = 2(n + 1)(2n + 1)

Finding a formula for the Area A

In this I am going to derive the formula for the area of a triangle algebraically. I know that the formula to find the area of a triangle is:

Area A =  × base× height

The area of the right-angled triangle with the Pythagorean triples of (2n + 1, 2n2 + 2n, 2n2 + 2n + 1) is obtained by substituting for the height and base (shortest and middle sides) from this set of triples as:

A =  (2n2 + 2n) (2n +1)

    =  2n (n +1) (2n + 1)

    = n(n + 1)(2n + 1)

The above formula can also be derived using the values of the area given in the sixth column of Table 1.

From the above I see that the third difference is constant and hence I will use the equations I have derived in Appendix C to find the nth term of the cubic sequence. Thus I find that:

6a = 12                (considering the 3rd difference)

  a = 2

12a + 2b = 30        (considering the 1st term of the 2nd difference)

Substituting for a in the above equation        ⇒ 24 + 2b = 30        ⇒ b = 3

7a + 3b + c = 24        (considering the 1st term of the 1st difference)

Substituting for a and b in the above equation

14 + 9 + c = 24        ⇒ c = 1

Considering the 1st term of the sequence S1 and substituting for a, b and c,

a + b + c + d = 6        ⇒ 2 + 3 + 1 + d = 6        ⇒  d = 0

So the formula to find the area A of the right-angled triangle with the Pythagorean triples of (2n + 1, 2n2 + 2n, 2n2 + 2n + 1) is:

A = an3 + bn2 + cn + d

   = 2n3 + 3n2 + n + 0

  = 2n3 + 3n2 + n

I can now factorise the RHS of this equation because I see that n is a common factor. Therefore it can now be written as:

A = n (2n2 + 3n+ 1)

The quadratic factor on the RHS can be further factorised as the product of two linear factors, as this is the simple quadratic expression, similar to which I have encountered before. Thus it can now be written as:

A = n (n + 1) (2n+ 1)

This is exactly the same formula I have derived earlier using algebra for the area of the right angle triangle.

At this stage it is worth incorporating my research on the number sequence. I found from an A-Level Core Mathematics book by Bostock and Chandler that the sum of the squares of the first n natural numbers Tn is given by the formula:

Tn =  n (n + 1) (2n + 1)

Thus it can be shown that the area A of the right angle triangle given above and Tn are related as:

A = 6Tn

Relationship between Perimeter and Area

In this I am going to show how the Perimeter and Area of a right angle triangle whose sides given by the Pythagorean triple of (2n + 1, 2n2 + 2n, 2n2 + 2n + 1) are related. It has been shown earlier that:

Perimeter P = 2(n + 1)(2n + 1)

Area A = n (n + 1) (2n + 1)

Dividing A by P, I find that

This can be easily verified, for example, for the 3rd Pythagorean triple n = 3, and from Table 1 P and A of this right angle triangle are 56 and 84 respectively, i.e. .

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Extensions to the Investigation

So far formulas have been derived for the nth Pythagorean triple (a, b, c) of a right-angled triangle with the assumption that all the side lengths of the triple are co-prime. I have proved that any Pythagorean triple of which the shortest side a is an odd number can be found using the formulas for (a, b, c) as:

( 2n + 1, 2n2 + 2n, 2n2 + 2n + 1)

Another suggestion to find a Pythagorean triple is to multiply a, b and c by any number.  This will then become the common factor of all the numbers of the triple, e.g. each number of the Pythagorean triple (3, 4, 5) can be multiplied by 2 to get (6, 8, 10), which is also a triple. For this triple 2 is a common factor.

Factorisation

The Pythagorean theorem states that:

a2 + b2  = c2

By rearranging the above, I get

b2 = c2 – a2

By using the identity x2 – y2 ≡ (x + y)(x – y), I can express

b2 = (c + a) (c – a)

If I choose a and c such that (c + a) (c – a) is a perfect square, then I have a Pythagorean triple.

 

So b2 is divisible by 4 since I am assuming that b is even, and both (c + a) and (c – a) are even since a and c are both odd. I can divide both sides by 4 and still be working with integer values, i.e.

Specific Example

The right side of the equation is still a perfect square, simply represented differently. The simplest way to get the left side to be a perfect square is to set (c + a) /2 and (c – a) / 2 integer values. For example, suppose (c + a) /2 = 52 and (c – a) / 2 = 42.

(i.e.) b2 / 4 = 25 × 16         ⇒ b2 = 4 × 16 × 25 = 1600 so b = 40

I now have two equations with two unknowns a and c:

Addition and subtraction of the above equations alternately yield:

c = 25 + 16 = 41

a = 25 – 16 = 9

This (9, 40, 41) has already been shown as a Pythagorean triple satisfying a2 + b2  = c2.

General Case

With appropriate choices of integers m and n (resemblance to a and c), I can get a Pythagorean triple by setting (c + a) / 2 = m2 and (c – a) / 2 = n2, and proceeding as above. In this it is assumed that m > n. I get:

                ⇒ b = 2mn

Adding and subtracting (c + a) / 2 = m2 and (c – a) / 2 = n2 alternately as I did above give:

a = m2 – n2

c = m2 + n2

So finally, I can say that any Pythagorean triple can be obtained by setting:

(a, b, c) = (m2 – n2, 2mn, m2 + n2)

There is a restriction that applies to all such triples produced in this way, i.e. m > n (otherwise I will get b as negative). Table 2 shows the sets of Pythagorean triples generated using different values of the integers m and n.

Multiples of Pythagorean Triples

Multiplying any Pythagorean triple by any positive integer yields another set of triples.

Specific example

Consider the first Pythagorean triple (3, 4, 5). Multiplying each side of the triple by 2 gives (6, 8, 10) which is also a Pythagorean triple as shown below.

Table 2 shows this to a greater extent. While considering multiples, it is interesting to refer back to the proof of showing that the formulas derived previously satisfy a2 + b2 = c2.

Effectively, in multiples (ma)2 + (mb)2 = (mc)2, should have the same effect, where m is any positive integer.

Proof

This is to show that by multiplying each side of a Pythagorean triplet by any positive integer will also produce another Pythagorean triple.

Shown in the above diagram are the sides of a right angle triangle with their sides (previously derived formulas in terms of the Pythagorean triplet number n) that are multiplied by a positive integer m.

It is thus sufficient to show that:

[m(2n + 1)]2 + [m(2n2 + 2n)]2 = [m(2n2 + 2n + 1)]2

Expanding and factorising the LHS give:

m2(2n +1)2 + m2(2n2 + 2n)2 = m2(2n2 + 2n + 1)2

m2[(2n +1)2 + (2n2 + 2n)2] = m2 (2n2 + 2n + 1)2 

Since I have proved that formulas for a2 and b2 sum up to c2 earlier, it is not needed to prove that m2 (2n2 + 2n + 1)2 equals m2[(2n +1)2 + m2(2n2 + 2n)2] as m2 is just a common factor on both sides of the equation.

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APPENDIX A

In this Appendix the nth term of a linear sequence is derived algebraically.

Let Sn be the nth term of a linear sequence given as:

Sn = an + b

where a and b are constants. Substitution of n = 1, 2, 3, 4 alternately in the above sequence gives:

S1 = a + b,        S2 = 2a + b,         S3 = 3a + b,        S4 = 4a + b and so on.

Equating the 1st difference and the 1st term to the corresponding values in any linear sequence, I can find the values of a and b, and hence the formula for the nth term of the linear sequence.

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APPENDIX B

In this Appendix the nth term of a quadratic sequence is derived algebraically.

Let Sn be the nth term of a quadratic sequence given as:

Sn =  an2 + bn + c

where a, b and c are constants. Substitution of n = 1, 2, 3, 4 alternately in the above sequence gives:

S1 = a + b + c,        S2 = 4a + 2b + c,        S3 = 9a + 3b + c

S4 = 16a + 4b + c and so on.

Noticing that the 1st difference is not constant, I went on to find the 2nd difference, and found that it is constant.

Equating the first terms of the 2nd and 1st differences, and the 1st term of Sn to the corresponding values in any quadratic sequence, I can find the values of a, b and c, and hence the formula for the nth term of the quadratic sequence.

APPENDIX C

In this Appendix the nth term of a cubic sequence is derived algebraically.

Let Sn be the nth term in a cubic sequence given as:

Sn = an3 + bn2 + cn + d

where a, b, c and d are constants. Substitution of n = 1, 2, 3, 4, 5 alternately in the above sequence gives:

Finding that the 1st and the 2nd differences are not constants, I went on to find the third difference, and found that this was constant.

Equating the first terms of the 3rd, 2nd and 1st differences, and the 1st term of Sn to the corresponding values in the sequence, I can find the values of a, b, c and d and hence the formula for the nth term of the cubic sequence.