For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

The simplest way to work this out would be to find a common difference between the numbers.

However 3 – 1 = 2

             5 – 2 = 3

             7 – 3 = 4

There is no common difference in the first difference so I will try the 2nd difference

3-2 = 1

4-3 = 1

This shows that there is a pattern between the numbers. So all I have to do now is find the formula. There might be a pattern if I multiply each number by the same number as in n².

However:

1²=1

2²=4

3²=9

This cannot be right.

Maybe it would work if I tried multiplying the numbers by 2 as in 2n.

However:

1x2=2

2x2=4

3x2=6

This is wrong. However I have noticed a pattern. Each of the numbers in this pattern is one less than the one I was aiming to get. This must mean that  

2n+1

is the correct formula   

To check this I will pick a random number between 1 and 25 and check it on the table above the n table. The number will be 25.

2x25+1=51.

This is the same as the number on the table so it must be right.

Next I will work out a formula to work out the middle side:

First I will see if there are any common differences.

There does seem to be a pattern because all the middle side numbers are multiples of 4.

12-4=8

24-12=12

12-8=4.

This again reinforces the fact that there is a pattern as the 2nd difference is 4. I will try squaring n to see if it gets to the number.

1²=1

2²=4

3²=9

These numbers are nowhere near the numbers I was trying to get. However they all have one thing in common and that is that they are all less than the numbers that I was trying to get. I will try multiplying the squared numbers by 2.

2 x 1² = 2

2 x 2² = 8

2 x 3² =18

I seem to be getting somewhere with this formula. I think that the first part of it will be 2n². I shall now try adding n to the formula to make the equation 2n² + n.

2 x 1² + 1 = 3

2 x 2² + 2 =10

2 x 3² + 3 = 21

I have noticed that each of these results is n less than the number I was aiming to get. This must mean that the formula is:

2n²+2n

I think this is the formula from looking at the first few terms but I will check it using another random number. The number will be 18.

2x18²+2x18=684

This is the same as the number given in the table so I presume that it is right.

I will now work out another formula to work out the longest side of the triangle using n:

This is probably the easiest to work out, as we know that the longest side is always one more than the middle side and that they are always consecutive numbers. So I can assume that the formula is:

2n²+2n+1

This makes sense but just to check I will use the number 6 to check the formula.

2x18²+2x18+1=685

This again fits in with the table above so it has to be right.

I am now going to see if there is a formula to work out the perimeter just using N.

The easiest thing to say would just be to work out the perimeter using the previous three equations. You could just say a + b + c. however if we join all the equations found already we should be able to get a unique formula to work out the equation. I will add the formula for the shortest side with the formulas for the middle side and longest side.

2n + 1 + 2n² + 2n + 2n² + 2n+1.

This formula simplified looks like this:

4n²+6n+2

This formula will work out the perimeter without having to know what the sides of the triangle are.

To check this I will use all three of the triples that I have been given

1. 4 x 1² + 6 x 1 + 2 = 12
2. 4 x 2² + 6 x 2 + 2 = 30
3. 4 x 3² + 6 x 3 + 2 = 50

All of these are correct and are as given in the table.

A formula to work out the area using N is what I need next.

I will see if there is any pattern between the numbers

Although there are probably patterns I have just noticed that all you have to do is multiply the equations for the three sides by each other. I will multiply the formula for the shortest side with the formulae for the middle and the longest side.

(2n+1)(2n²+2n)(2n²+2n+1)

Simplified this equation looks like this:

2n³+3n²+n

To check this I will again use all the three numbers in the table.

1. 2 x 1³ + 3 x 1² + 1 = 6
2. 2 x 2³ + 3 x 2² + 2 = 30
3. 2 x 3³ + 3 x 3² + 3 = 84

These are all correct.

So, the formulae that I have worked out are:

• Shortest side=2n+1
• Middle side=2n²+2n
• Longest side=2n²+2n+1
• Perimeter=4n²+6n+2
• Area=2n³+3n²+n

These are all the formulae that you can work out using the n. this means that you can work out any number in the sequence. It is different from the other formula that I worked out before this because that was working out the sides of the triangle using the shortest side. Each formula is just as effective.

I will now put all the numbers that I have worked out into a table up to the 25th term.

Another pattern that I have noticed is that there must be a connection between the sides of the triangle because the middle side and the longest side have a difference of one. I will now try and work out the formula. I will basically use trial and error and try out different formulae that I think of.

(Shortest side) ²+(middle side) ²=(longest side) ²

I already know that this is correct because it is Pythagoras theorem. However I will still check it using the Pythagorean triple 3,4,5.

3²+4²=5²

I know that this is correct because

3²=9

4²=16

9 + 16 = 25, which is also 5²

(Middle side) ²+(longest side) ²=(shortest side) ²

To see if this statement is true I shall use the Pythagorean triple 9,40,41.

40²+41²=9²

This is not true because

40² = 1600

41² = 1681

which add together to make 3281

and

9²=81

This means that this is not a correct formula.

(Longest side) ²/(shortest side) ² = (middle side)

To check this I will use the Pythagorean triple 15,112,113.

113²/15²=112²

This is wrong as

113²=12769

15²=225

and

12769/225 = 56.751111

and as

The middle side = 113 this formula has to be wrong.

Shortest side x longest side = (middle side) ²

To check this I will use the Pythagorean triple 7,24,25

7 x 25 = 175

This is wrong because

24²= 576 and this is very different

(Longest side) ²-(middle side) ²=shortest side

To check this formula I will use the Pythagorean triple 5,12,13

13²-12²=5²

This is right as

13²=169

12²=144

169-144=25 which is also 5².

So this formula is correct

(Middle side) ²+Smallest side=Largest side

To check this I will use the Pythagorean triple 5, 12,13.

12²+5=13

This is silly and I shouldn’t really have tried it, as I should have been able to work out that this formula is incorrect just by looking at the numbers involved. 144 is bigger than 13 without adding the 5.

Largest side + Middle side = (shortest side) ²

To check this I will use the Pythagorean triple 3,4,5

5+4=3²

This is correct because 5 + 4 = 9 which is also 9.

Just to check that this formula works with all triples I shall pick 3 random numbers from the N column of the table and use the formula with them. I shall use (i) 23 (ii) 10 (iii) 16

1. 1105+1104=47² which is also 2209. This is correct
2. 221+220=21² which is also 441. This is correct
3. 545+544=33² which is also 1089. This is correct

This seems to be a correct formula.

So the formulae  I have worked  out in this section are:

• (Shortest side)²+(middle side)²=(longest side)²
• (Longest side) ²-(middle side) ²=shortest side
• Largest side + Middle side = (shortest side) ²

All these are based on the assumption that the difference between the longest side and the middle side is 1.

However, I am interested about what happens when the difference between the longest side and the middle side is different from 1. I shall investigate this going up to c=b+10.

In the next few investigations I was required to work out the square roots of numbers. However, although it is possible to work out a square root of every number I chose only to investigate the perfect squares. Those numbers which are whole numbers. This is because I am only investigating Pythagorean triples with positive integers.

Investigating c=b+2

I will now investigate when longest side – middle side= 2. (c-b=2)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 2(2b+2)

   =4b+4

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+3

I will now investigate when longest side – middle side=3. (c-b=3)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 3(2b+3)

   =6b+9

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+4

I will now investigate when longest side – middle side=4. (c-b=4)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 4(2b+4)

   =8b+16

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+5

I will now investigate when longest side – middle side=5. (c-b=5)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 5(2b+5)

   =10b+25

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+6

I will now investigate when longest side – middle side=6. (c-b=6)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 6(2b+6)

   =12b+36

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+7

I will now investigate when longest side – middle side=7. (c-b=7)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 7(2b+7)

   =14b+49

I can now produce a table for different values of b. Perfect squares are coloured red.

There are no triples that have all whole numbers in this formula.

Investigating c=b+8

I will now investigate when longest side – middle side=8. (c-b=8)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 8(2b+8)

   =16b+64

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+9

I will now investigate when longest side – middle side=9. (c-b=9)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 9(2b+9)

   =18b+81

I can now produce a table for different values of b. Perfect squares are coloured red.

Investigating c=b+10

I will now investigate when longest side – middle side=10. (c-b=10)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 10(2b+10)

   =20b+100

I can now produce a table for different values of b. Perfect squares are coloured red.

EXTENSION

Up to now I have worked out all these formulae and sequences by myself. However I wondered how many other formulae had been discovered and how other people worked out these sequences. This resulted in me looking in encyclopaedias and on the Internet to see what other people had to say. In each of these cases I have written an account in my own words of what each place said to do and then I have given an example worked out by myself.

Encyclopaedia Britannica

An interesting discovery which I made when looking through the Encyclopaedia Britannica was this:

They had worked out a formula for working out the three sides of a triangle.

They say that if you choose 2 numbers that are neither both even nor both odd, and that are both relatively prime (this means that they share no common factor apart from 1), then make the larger number p and the smaller number q.

If you do this then there are 3 formulae that can work out the sides of a triangle for you. They are:

1. p²-q²
2. 2pq
3. p²+q²

I shall try this using a table and 5 randomly chosen pairs of numbers

I know that the first 3 are correct as they are the same answers as those given on page 8. however I shall check that the last 2 are correct by seeing if they fit in with the formula a²+b²=c².

47²=2209

1104²=1218816

1105²=1221025

2209+1218816=1221025

This means that the formula was correct and worked out this triple correctly.

I shall now check the final triple,

10185²=103734225

808²=652864

10217²=104387089

103734225+652864=104387089

This means that the formula was correct and worked out this triple correctly.

Even though this is a good formula I have worked out from further research that if you use this formula it will not generate every single triple.

Internet

An interesting thing I found on the Internet, which I had never heard of before, was Pythagorean Quads (quadruples).

Pythagorean quads are the 3D version of the triples and can all be generated by the following formula, where a, b, c and d are whole numbers.
If p = a2 + b2 - c2 - d2

   q = 2(ad - bc)  

   r = 2(ac + bd)

and s = a2 + b2 + c2 + d2 

then p2 + q2 + r2 = s2.

I shall now try and work out a few of these to check if this formula is correct

I shall choose the numbers a=4, b=1, c=1 and d=1.

This means that p=16+1-1-1 = 15

                      q=2(4-1)=6

                       r=2(4+1)=10             

                       s=16+1+1+1=19

From looking further on the Internet for confirmation I conclude that this formula does work because the quad that I worked out above is correct.

        

Conclusion

 I started this investigation by noticing that there was a difference of 1 between the middle side and the longest side. I then went on to work out formulae relating to this. All these formulae were based on the assumption that c-b+1. however after investigating further I realised that the difference between the middle side and the longest side could be any number. So I investigated all the differences up to a difference of 5. I decided that I would stop each table at the value of b=25. I realised that this investigation could go on forever and you could investigate unlimited values of b and  there are unlimited differences that I could have investigated but I had to set a limit. I also decided that I would only investigate up to c=b+10 as I felt that was a sufficient number of investigations. I wrote the extension because I thought it was interesting to see what else other people had discovered