Beyond Pythagoras

From no.5 the sequence = 11, so for no.6 the sequence will = 13, for no.7 the sequence will = 15, for no.8 the sequence will = 17, for no.9 the sequence will = 19, finally for no. the sequence will = 21

I was correct!!!! The formula does work.

Now I’ve worked out a successful formula for the small side of family 1 I will follow the same methods I used to work out formulas for the middle & longest sides of family 1.

Middle

From looking at this table I worked out that the difference between each sequence goes up by +4 every time from what you added to get your previous sequence. I also found that if you use the 2nd difference (the amount you always add on to your last total which gave you your previous number in the sequence) and half it (2 in this case) then add that behind an n2. So now I’ve got 2n2 now if I add the sequence from imaginary term 0 I get 2n2 + 0. Finally if I substitute 1 into the formula I’ve got and do this for the 1st term you get 2, now I just need to put this behind an n (2n). So if I add this onto the end of the 2n2 I got earlier I get 2n + 2n2 which I believe is the formula for this side of family 1!!!!!!! This method works when working out any nth terms which change in difference between each sequence.

The formula for the medium side = 2n + 2n2 

I predict that for no. 7 the sequence will be (2x7) + (2x72) = 112

Now I’ll prove I’m right:

Between each sequence you add on an extra 4 from what you added to get the previous sequence so no.5 =60, no.6 =84, no.7 = 112, again I was correct!!!!

Ok I’ve successfully worked out a formula for the smallest side. I have also worked out a formula for the middle side!!! Now I need to work out a successful formula for the largest side of family 1’s Pythagorean triples!!!!

Large

I will use the same method I did last time for the middle side of family 1: (2n2) + (1) + (2n) = 2n + 2n2 + 1

I have concluded that the formula = 2n + 2n2 + 1

I predict that for no.9 the sequence will be (2x9) + (2x92) + 1 = 181

Now again I’ll prove I’m right:

Between each sequence you add on an extra 4 from what you added to get the previous sequence so no.5 = 61, no.6 = 85, no.7 = 113, no.8 = 145, no.9 = 181

BINGO!!!! I was correct again!!!!!!

Now I have a formula for each side of family 1 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 + 1

b (middle side) = 2n + 2n2 

c (largest side) = 2n + 2n2 + 1

a2 = (2n+1) x (2n+1)

= 4n2 + 1 + 2n + 2n

= 1 + 4n + 4n2

b2 = (2n + 2n2) x (2n + 2n2)

= 4n2 + 4n4 + 4n3 + 4n3

= 4n2 + 8n3 + 4n4

c2 = (2n + 2n2 + 1) x (2n + 2n2 + 1)

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

= 1 + 4n + 8n2 + 8n3 + 4n4

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 1 are therefore correct!!!!

so……      

c2 = 1 + 4n + 8n2 + 8n3 + 4n4

a2         (1 + 4n + 4n2)

b2          (4n2 + 8n3 + 4n4)

(1 + 4n + 4n2) + (4n2 + 8n3 + 4n4)

= 1 + 4n + 8n2 + 8n3 + 4n4

My formulas for family 1 suit the equation a2 + b2 = c2, therefore they are correct!!!!!

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

I’m told that if I’m to generate family 2:

●I must double family 1 to find every other triple of family two

●the smallest side is even

●the difference between the middle and longest term is + 2

●find middle term pattern to generate series

Below is a table of all terms for family 2:

                                                                                       

                                                                                       

Family 2

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

Above you can see in the table for family 3 I have again added notes on patterns to the table, as I have for all the families so far. You may be thinking how do I know all these, triples, and how do I know that there are no missing triples? Well there’s a simple answer to that, notice that there is a gap between the small term of triples for family 3 of 6, well if you think there is a missing gap you would half the six to three because it’s lap bang in the middle, this is the same for sequences and triples, ok so if I try to add 3 on to the previous sequence of a small term number notice you always end up with an even number, this therefore breaks the rule of family 3 meaning there are no missing triples. All I have to do for family 3 is work out the formulas, like I did for family 1.

Small

Again using the same method for working out the nth term of something which always has the same difference between each sequence: (6n) + (3) = 6n + 3

The formula = 6n + 3

I predict that for no.6 the sequence will be (6x6) + 3 = 39

Now again I’ll prove I’m right:

Between each sequence you add on an extra 6 so no.5 = 33, no.6 = 39

I was correct the formula works!!!!!

Middle

The formula = 6n + 6n2 

I predict that no.7 will be (6x7) + (6x72) = 336

Between each sequence you add on an extra 12 to what you added to get the previous sequence so no.5 = 180, no.6 = 252, no.7 =336

Again I was correct!!!!!!

Large

I will use the same method I did last time for the middle side of family 1: (6n2) + (3) + (6n) = 6n + 6n2 + 3

The formula = 6n + 6n2 + 3

I predict that no.11 will be (6x11) + (6x112) + 3 = 795

Between each sequence you add on an extra 12 to what you added to get the previous sequence so no.5 = 183, no.6 = 255, no.7 = 339, no.8 = 435, no.9 = 543, no.10 = 663, no.11 = 795

Again I 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) = 6n + 3

b (medium side) = 6n + 6n2

c (largest side) = 6n + 6n2 + 3

a2 = (6n + 3) x (6n + 3)

= 36n2 + 9 + 18n + 18n

= 9 + 36n + 36n2   

b2 = (6n + 6n2) x (6n + 6n2)

= 36n2 + 36n4 + 36n3 + 6n3

= 36n2 + 72n3 + 36n4 

c2 = (6n + 6n2 + 3) x (6n + 6n2 + 3)

= 36n2 + 36n3 + 18n + 36n3 + 36n4 + 18n2 +18n + 18n2 + 9

= 9 + 36n + 72n2 + 72n3 + 36n4

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 3 are correct!!!!

so....

c2 = 9 + 36n + 72n2 + 72n3 + 36n4

a2        (9 + 36n + 36n2)   

b2        (36n2 + 72n3 + 36n4)

= (9 + 36n + 36n2) + (36n2 + 72n3 + 36n4) 

= 9 + 36n + 72n2 + 72n3 + 36n4

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

To summarise I have now got 3 formulas for each family I explored of Pythagorean triples.

I’m now going to move onto general formulas, which will be in terms of a, also I will have one formula for each family.

Family 1

a, b,  c= b +1

a = a

b =?

c2 = a2 + b2

(b+1) x (b+1) = b2 + 1 + b + b

So: 1 + 2b + b2 = a2 + b2

So if I switch the -1 below for the b2 I can get the b out of the formula……

b2 + 2b – b2 + a2 – 1

so……

b= a2 – 1        , so if b = a2 – 1     , then c =   a2 – 1      

        2                            2                            2

                                                                But you can’t have +1 on    

                                                                the end so we can times

                                 2c = a2 + 1                the formula by 2

                                   

        So…

             

Family 2

a, b, c = b + 2

a = a

b = ?

c2 = a2 + b2

(b+2) x (b+2) = b2 + 4 + 2b + 2b

So: 4 + 4b + b2 = a2 + b2

So if I switch the -4 below for the b2 I can get the b out of the formula……

b2 + 4b – b2 = a2 – 4

so……

b = a2 – 4

         4               so if b = a2 – 4       , then c = a2 - 4

                                           4                             4

But you can’t have +2 on the end so we can times the formula by 4

                                             4c =a2 -4 + 8

                                                    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 -12          c = 32 + 12

                  2 x 1                  2 x 1

That all works!!!!!!!

I’ll use another example family 2:

a = 6     b = 62 - 22             c = 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.