Beyond Pythagoras

In this piece of coursework I will be exploring Pythagorean triples which are beyond Pythagoras.

Pythagoras is where you have a right angled triangle and you know the values of sides a + b but you don’t know what side c is (hypotenuse). To calculate the hypotenuse you can square sides a + b and add the two answers together as one total. This total is equal to c2 so all you have to do now is find the square root of the total and you have worked out side c.

A Pythagorean triple is where the first two sides of a triangle (a + b) suit this equation: a2 + b2 = c2. For example lets say a=1, b=2, c=3. Now let’s try this in a Pythagoras equation: a2 + b2 = c2 (12 + 22 = 32) this is not correct!!! 12 + 22 = 5 we should know that 32 = 9!!!! From this we should acknowledge that the 3 numbers I used don’t fit the equation a2 + b2 = c2, this therefore means they are not Pythagorean triples!!!!

So from this you should notice that a2 + b2 = c2, if a=3, b=4, and c=5 (32+44 = 52) we can see that 32 + 42= 52(25) this therefore means these numbers make a Pythagorean triple.

I have been given 3 sets of Pythagorean triples from family 1 to analyse:

I’ve been told that family 1 of the Pythagorean triples has the following features:

●smallest side is odd

●the longest side is one more than the middle side

●on the middle side you add 4 more on than the last time

If I wanted to work out more Pythagorean triples for family 1 I can search for patterns from the data I have been provided with.

S

M

L

6

8

10

? (8)

? (15)

? (17)

10

24

26

? (12)

? (35)

? (37)

14

48

50

? (16)

? (63)

? (65)

Notice that I have put question marks in-between triples. This is because there are a set of hidden triples in-between the triples I already know. There are a set of hidden triples in family 2 because you all of a sudden notice that you can add another set of triples into the family 2 sequences and the smallest side still is even, the middle and longest side still has a difference of 2

I worked out easily that on the smallest term that the pattern between the triples is + 2 this is because this is the only way all the smallest sides of a triple in the sequence would be even.

Small

I will use the method I used before for working out the nth term of something which always has the same amount of difference between each sequence: (2n) + (4) = 2n + 4

The formula = 2n + 4

I predict that for no.6 the sequence will be (2x6) + 4 = 16

Now again I’ll prove I’m right:

Between each sequence you add on an extra 2 so no.5 = 14, no.6 = 16

The formula works, therefore I am correct!!!

The middle side was more difficult to work out!!! I did not spot any patterns until I worked out the formula and added the missing values!!! But when I did add the missing sequences I found:

Middle

I will use the same method I did last time for the middle side of family 1: (1n2) + (3) + (4n) = 4n + 1n2 + 3

The formula = 4n + n2 + 3

I predict that no.8 will have (4x8) + (82) + 3 = 99

Between each sequence you add on an extra 2 from what you added to get the previous sequence so no.5 = 48, no.6 = 63, no.7 = 80, no.8 = 99

I am correct, the formula works!!!!!

Large

Working out the formula for the largest term of family 2 was pretty easy, I worked it out straight away because it follows the same pattern as the middle term for family 2 the only difference between the two formulas is that you add different values at the end of the formula. I didn’t need to follow my method because I guessed straight away that the sequence was similar to that of the middle side of family 2.

The formula = 4n + n2 + 5

To prove my formula is correct I predict with my formula that no.10 will be (4x10) + (102) + 5 = 145

Between each sequence you add on an extra 2 from what you added to get the previous sequence so no.5 = 50, no.6 = 65, no.7 = 82, no.8 =101, no.9 =122, no.10 = 145

The formula was correct!!!!!

Now I have a formula for each side of family 2 I can test them by substituting into Pythagoras (a2 + b2 = c2) If the formulas are correct they will suit the Pythagoras equation.

a (smallest side) = 2n + 4

b (medium side) = 4n + n2 + 3

c (largest side) = 4n + n2 + 5

a2 = (2n + 4) x (2n + 4)

= 4n2 + 16 + 8n + 8n

= 16 + 16n + 4n2

b2 = (4n + n2 + 3) x (4n + n2 + 3)

= 16n2 + 4n3 + 12n + 4n3 + n4 +3n2 + 12n + 3n2 + 9

= 9 + 24n + 22n2 + 8n3 + n4

c2 = (4n + n2 + 5) x (4n + n2 + 5)

= 16n2 + 4n3 + 20n + 4n3 + n4 + 5n2 + 20n +5n2 + 25

= 25 + 40n + 26n2 + 8n3 + n4

So now I have put the formulas into a Pythagoras equation I need to check if a2 + b2 = c2, if it does then my formulas for family 2 are correct!!!!

so....

c2 = 25 + 40n + 26n2 + 8n3 + n4

a2         (16 + 16n + 4n2)

b2          (9 + 24n + 22n2 + 8n3 + n4)

(16 + 16n + 4n2) + (9 + 24n + 22n2 + 8n3 + n4)

= 25 + 40n + 26n2 + 8n3 + n4

My formulas for family 2 suit the Pythagoras equation, therefore they are correct!!!!

Now I have finished working on family 1 & 2, I will explore family 3........

I’m told that if I’m to generate family 3:

●I must triple family 1 to find family 3

●the smallest side is odd

●the difference between the middle and longest term is + 3

●find middle term pattern to generate series

Family 3

                                                    so…..

Family 3

a, b, c = b + 3

a = a

b = ?

c2 = a2 + b2

(b+3) x (b+3) = b2 + 9 + 3b + 3b

So: b2 + 6b +9 = a2 + b2

So if I switch the -9 below for the b2 I can get the b out of the formula……

b2 + 6b – b2 = a2 - 9

so………

b = a2 - 9

         6                 so if b = a2 – 9       , then c = a - 9

         6                             6

But you can’t have +3 on the end so we can times the formula by 6

                                                                  6c = a2 + 9

                                                            so………

Now I have all the general formulas I will look to see if I can combine all the formulas together so I can find a formula for each side.

Side a = a

Side b = a2 – x2

2x

Before I worked out the formula for side b I noticed some patterns, let’s call the families x: the denominator of the formula for side b is always 2x, and then I noticed on top after the a2 it’s always x2!!!!

Side c = a2 + x2

2x

Side c was very straight forward to work out the only thing that changes from the side b formula is the sign which changes to a +!!!

Now from the formulas I have substituting families as x I can work out any triples if given side a to begin with!!!1

I’ll prove I’m right:

I’ll use family 1:

a = 3     b = 32 -12c = 32 + 12

                  2 x 1                  2 x 1

That all works!!!!!!!

I’ll use another example family 2:

a = 6     b = 62 - 22c = 62 + 22

                   2 x 2                    2 x 2

This also works!!!!!

Finally I’ll try out family 3:

a = 9     b = 92 - 32 c = 92 + 32

                   3 x 3                    3 x 3

I’ve successfully worked out a general formula for each side of any family, this means I could work out any triples I want in any sequences in family 1, 2, or 3. I could also check to see if a set of Pythagorean triples is in any of the first 3 families.