### 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 timeIf I wanted to work out more Pythagorean triples for family 1 I can search for patterns from the data I have been provided with.The easiest way to work out if there is a pattern is to find a formula which will enable you to work out any part of the sequence!As you can see in the table below there are 5 sets of Pythagorean triples for family 1. I was given 3 Pythagorean triples to begin with from family 1 but I have worked out and added two additional Pythagorean triples to the table (which are in green)Family 1SML345512137242594041116061138485As you can see I have spotted patterns in family 1 I should be able to find a formula if I separate each side of family 1.

I will make tables for the sequences of each side of any family, or patterns when I’m ever trying to find the nth term of anything.SmallFrom looking at this table I have noticed that there is a pattern between how much you how much you add to get to the next sequence, in this case it’s always +2. Also when working out this formula I noticed that the difference between each sequence which in this sequence is repeatedly +2 goes before n in the final formula. So I have 2n also notice that by working out the imaginary 0th term I can add the sequence number from that term and add this to the 2n, it looks like I’ve just successfully worked out a formula!!! This method works when working out the nth term for something which always uses the same figure to get to the next sequence.I worked out that the formula = 2n + 1I will make a prediction to prove my formula works:I predict that for no.

10 the sequence will be (10×2) + 1 = 21.How can I prove I’m right?Looking for patterns in the sequence helps, for instance in this table for the small side of family 1 I noticed that you add two on every time you move up the sequence. I’ll test this against what my formula predicts: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 = 21I 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.MiddleFrom 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 + 2n2I predict that for no. 7 the sequence will be (2×7) + (2×72) = 112Now 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!!!!LargeI will use the same method I did last time for the middle side of family 1: (2n2) + (1) + (2n) = 2n + 2n2 + 1I have concluded that the formula = 2n + 2n2 + 1I predict that for no.9 the sequence will be (2×9) + (2×92) + 1 = 181Now 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 = 181BINGO!!!! 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 + 1b (middle side) = 2n + 2n2c (largest side) = 2n + 2n2 + 1a2 = (2n+1) x (2n+1)= 4n2 + 1 + 2n + 2n= 1 + 4n + 4n2b2 = (2n + 2n2) x (2n + 2n2)= 4n2 + 4n4 + 4n3 + 4n3= 4n2 + 8n3 + 4n4c2 = (2n + 2n2 + 1) x (2n + 2n2 + 1)= 4n2 + 4n3 + 2n + 4n3 + 4n4 + 2n2 + 2n + 2n2 + 1= 1 + 4n + 8n2 + 8n3 + 4n4So 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 + 4n4a2 (1 + 4n + 4n2)b2 (4n2 + 8n3 + 4n4)(1 + 4n + 4n2) + (4n2 + 8n3 + 4n4)= 1 + 4n + 8n2 + 8n3 + 4n4My 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.

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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 seriesBelow is a table of all terms for family 2:Family 2SML6810? (8)? (15)? (17)102426? (12)? (35)? (37)144850? (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 2I 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.SmallI 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 + 4The formula = 2n + 4I predict that for no.6 the sequence will be (2×6) + 4 = 16Now again I’ll prove I’m right:Between each sequence you add on an extra 2 so no.

5 = 14, no.6 = 16The 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:MiddleI will use the same method I did last time for the middle side of family 1: (1n2) + (3) + (4n) = 4n + 1n2 + 3The formula = 4n + n2 + 3I predict that no.8 will have (4×8) + (82) + 3 = 99Between 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 = 99I am correct, the formula works!!!!!LargeWorking 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 + 5To prove my formula is correct I predict with my formula that no.

10 will be (4×10) + (102) + 5 = 145Between 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 = 145The 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 + 4b (medium side) = 4n + n2 + 3c (largest side) = 4n + n2 + 5a2 = (2n + 4) x (2n + 4)= 4n2 + 16 + 8n + 8n= 16 + 16n + 4n2b2 = (4n + n2 + 3) x (4n + n2 + 3)= 16n2 + 4n3 + 12n + 4n3 + n4 +3n2 + 12n + 3n2 + 9= 9 + 24n + 22n2 + 8n3 + n4c2 = (4n + n2 + 5) x (4n + n2 + 5)= 16n2 + 4n3 + 20n + 4n3 + n4 + 5n2 + 20n +5n2 + 25= 25 + 40n + 26n2 + 8n3 + n4So 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 + n4a2 (16 + 16n + 4n2)b2 (9 + 24n + 22n2 + 8n3 + n4)(16 + 16n + 4n2) + (9 + 24n + 22n2 + 8n3 + n4)= 25 + 40n + 26n2 + 8n3 + n4My 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…

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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 seriesFamily 3SML9121515363921727527120123331801833925225545336339Above 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.

SmallAgain using the same method for working out the nth term of something which always has the same difference between each sequence: (6n) + (3) = 6n + 3The formula = 6n + 3I predict that for no.6 the sequence will be (6×6) + 3 = 39Now again I’ll prove I’m right:Between each sequence you add on an extra 6 so no.5 = 33, no.6 = 39I was correct the formula works!!!!!MiddleThe formula = 6n + 6n2I predict that no.7 will be (6×7) + (6×72) = 336Between 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 =336Again I was correct!!!!!!LargeI will use the same method I did last time for the middle side of family 1: (6n2) + (3) + (6n) = 6n + 6n2 + 3The formula = 6n + 6n2 + 3I predict that no.11 will be (6×11) + (6×112) + 3 = 795Between 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 = 795Again 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 + 3b (medium side) = 6n + 6n2c (largest side) = 6n + 6n2 + 3a2 = (6n + 3) x (6n + 3)= 36n2 + 9 + 18n + 18n= 9 + 36n + 36n2b2 = (6n + 6n2) x (6n + 6n2)= 36n2 + 36n4 + 36n3 + 6n3= 36n2 + 72n3 + 36n4c2 = (6n + 6n2 + 3) x (6n + 6n2 + 3)= 36n2 + 36n3 + 18n + 36n3 + 36n4 + 18n2 +18n + 18n2 + 9= 9 + 36n + 72n2 + 72n3 + 36n4So 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 + 36n4a2 (9 + 36n + 36n2)b2 (36n2 + 72n3 + 36n4)= (9 + 36n + 36n2) + (36n2 + 72n3 + 36n4)= 9 + 36n + 72n2 + 72n3 + 36n4My 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 1a, b, c= b +1a = ab =?c2 = a2 + b2(b+1) x (b+1) = b2 + 1 + b + bSo: 1 + 2b + b2 = a2 + b2So if I switch the -1 below for the b2 I can get the b out of the formula…

…b2 + 2b – b2 + a2 – 1so….

..b= a2 – 1 , so if b = a2 – 1 , then c = a2 – 12 2 2But you can’t have +1 onthe end so we can times2c = a2 + 1 the formula by 2So…Family 2a, b, c = b + 2a = ab = ?c2 = a2 + b2(b+2) x (b+2) = b2 + 4 + 2b + 2bSo: 4 + 4b + b2 = a2 + b2So if I switch the -4 below for the b2 I can get the b out of the formula.

…..b2 + 4b – b2 = a2 – 4so.

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..b = a2 – 44 so if b = a2 – 4 , then c = a2 – 44 4But you can’t have +2 on the end so we can times the formula by 44c =a2 -4 + 8so..

…Family 3a, b, c = b + 3a = ab = ?c2 = a2 + b2(b+3) x (b+3) = b2 + 9 + 3b + 3bSo: b2 + 6b +9 = a2 + b2So if I switch the -9 below for the b2 I can get the b out of the formula…

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b2 + 6b – b2 = a2 – 9so…..

….b = a2 – 96 so if b = a2 – 9 , then c = a – 96 6But you can’t have +3 on the end so we can times the formula by 66c = a2 + 9so………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 = aSide b = a2 – x22xBefore 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 + x22xSide 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!!!1I’ll prove I’m right:I’ll use family 1:a = 3 b = 32 -12 c = 32 + 122 x 1 2 x 1That all works!!!!!!!I’ll use another example family 2:a = 6 b = 62 – 22 c = 62 + 222 x 2 2 x 2This also works!!!!!Finally I’ll try out family 3:a = 9 b = 92 – 32 c = 92 + 323 x 3 3 x 3I’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.