From the table I am able to notice various patterns:
- Both the smallest and largest numbers are odd
- The small side numbers go up by 2 each time
- All middle numbers are even
- The middle numbers are multiples of 4
- The largest numbers are 1 number larger than the middle
- The area and perimeter are always even
I can also see a formula to get the smallest side:
2n + 1 = S (S = Smallest Side, n = Term Number)
This formula will work with all Pythagorean triples.
I will use T3 and T4 to prove this:
Term 3 Term4
2n + 1 = S 2n + 1 = S
2 x 3 + 1 = 7 2 x 4 + 1 = 9
As you can see the formula works.
A question now comes to mind, are there any triples which do not fit an odd, even, odd pattern?
The way in which I can see is to try and investigate this by finding a pattern for the M (middle) side.
By looking at the chart which I created, I can see that there is some sort of pattern as it shows that the middle sides are multiples of 4.
If I use the method of difference to look at the second difference between this set of numbers, I can then see the if there are any quadratic functions, then use them to identify them.
With the repetition of 4, I notice there is a quadratic function going on.
Quadratic functions look like the following:
an² + bn + c
Getting the square root of the second difference (4) I will then be left with “a” the co-efficient of “n²”.
2n²=4
I am now able to fine “c” by calculating the output for T0. If I subtract 4 from the first difference between the 1st and 2nd term and then subtract the output of the first term from my resulting number, I am given “c”.
I know this works because when n = 0, an² + bn also is 0, this then means that “c” is the only number contributing to the answer.
e.g.
8 – 4 = 4
4 – 4 = 0
c = 0
Now I can calculate “b” by substituting “a” and “c” for the values which have now been found. I will use T1, that way I can find the un-multiplied value of “b”.
an² + bn + c = 4
2(1²) + b(1) + 0 = 4
2 + b + 0 = 4
-2 -2
b = 4-2
b = 2
This shows that “a” = 2 “b” = 2 “c” = 0
I can now create a formula which finds the middle length from the term number.
T_ 2 x n² + 2n = M or 2n² + 2n = M
I will prove this works by testing it on a triangle which I know the M term number for.
The triangle on the T3 (3rd term) has a M value of 24:
2n² + 2n = M 2 x 3² + 2 x 3 = M
2(3m²) + 2 (3) = 24 or 18 + 6 = 24
18 + 6 = 24
The formula works.
I am now going to look at using the method to find a formula for the longest side (L).
The L is always one extra unit than the M value. So I presume that the formula will be the same but adds an extra unit. Because it is adding 1 consecutive unit each time, it must be “c” value which equals 1, or the gap would increase each time the term number rose.
I predict L = M + 1, or 2n² + 2n + 1 = L
I will now test the prediction
8 – 4 = 4
5 – 4 = 1
c = 1
2n² + 2n + 1 = L
This means my prediction was correct.
I will use T4 to prove this:
2n² + 2n + 1 = L
2 x (4²) + 2 (4) + 1 = 41
32 + 8 + 1 = 41
I am now wondering if I can find a formula to find the perimeter.
I shall use the algebraic formula to try and find a formula for finding the perimeter by using the term number.
Perimeter will = P
P = S + M + L
(2n + 1) + (2n² + 2n) + (2n² + 2n + 1) = P
This can be simplified to
4n² + 6n + 2 = P
Now I’ll use T4 (term 4), and input it into the formula to see if it works.
4n² + 6n + 2 = P
(4 x 4²) + (6 x 4) + 2 = P
(4 x 16) + 24 + 2 = P
90 = P
The length of the sides for T4 is 9 (S) , 40 (M) , and 41 (L)
S + M + L = P
9 + 40 + 41 = 90
This proves the formula works.
I will now try and use the same formula but modify it in some ways to see if I can find a formula for area.
Area = A
A = 0.5 x S x M (formula for a triangle)
(2n + 1) x (2n² + 2n)
2n (2n² + 2n) + 1 (2n² + 2n)
4n³ + 4n² + 2n² + 2n
4n³ + 6n² + 2n
I’ll now test the formula using T3:
(4 x 3³) + (6 x 3²) + (2 x 3)
(4 x 27) + (6 x 9) + 6
108 + 54 + 6
The lengths of the sides for T3 are 7, 24, 25
7 x 24 168
This proves the formula is correct.
Part 2:
I am now going to investigate Pythagorean triples which have S as an even number.
The question which I am curious is, will the same rules apply?
S = 6
This triangle is the first odd triangle, so by re-arranging the formula to find S I am then able to find if the formula works. I should get Tn = 1
2n + 1 = S
2n + 1 = 6
- 1 = 6 – 1 2n = 5
n = 5/2
n = 2.5
This proves that the formula does not work.
Does the triangle have any relationship with the odd number T1?
T S M L
3 4 5
6 8 10
Things which I have noticed:
- Odd 5 – 4 = 1, even 10 – 8 = 2
- Triangles have the same angles (right angle)
- The sides of the even triangle are double the sides of the odd i.e. 3, 4, 5 and 6, 8, 10
-
They are similar to the ratio 1:2 3, 4, 5
6, 8, 10
I am now curious as to whether the perimeter and area will also double.
The perimeter doubles (like the difference in M and L) but the area gets multiplied by 4.
So now I have to add “x4” and “x2” to my existing formulas.
A = 4 (0.5 x S x M)
A = 4 [2n(2n² +2n) + 1(2n² + 2n)]
2
A = 4
P = 2(4n² + 6n + 2)
P = 8n² + 12n + 4
However the formula for the side lengths also needs to be modified.
I think I will be able to just multiply my existing formula by the two lengths that are in a ratio of 1:2. That would give me the following:
S = 2 (2n + 1)
S = 4n + 2
M = 2 (2n² + 2n)
M = 4n² + 4n
L = 2 (2n² + 2n + 1)
L = 4n² + 4n + 2
Using T2, the next triple with an even S would then be:
S = 4 x 2 + 2 = 10 S M L
T1 3 4 5
M = 4 x 2² + (4 x 2) = 24 6 8 10
T2 5 12 13
L = 4 x 2² + (4 x 2) + 2 = 26 10 24 26
These are also double the lengths of the odds of the T2 triple (5, 12, 13)
I have noticed that:
- All the side lengths are even because they are multiples of two.
- 2 is the difference between the even M and L sides.
Would it then mean that, for Pythagorean where the difference between the M and L sides is 3 that the ratio would be 1:3, which would then make the formulas for length and perimeter, be multiplied by 3 also?
Now thinking about the area, which is the S and M sides multiplied by each other and then divided by 2.
3S x 3M x 0.5, can be the same as:
or
The question is, would the area be the same formula multiplied by 9?
The even triangle which I was looking at was 2S x 2M x 0.5 =
2(2n + 1) x 2 (2n² + 2n) x 0.5
which can also be written
4x(2n + 1) x (2n² + 2n)
I will use the ratio 3:1 to expand the odd T1. Which means 3, 4, 5 then becomes
9, 12, 15.
The triangle is a right angled triangle, I am curious to see if my formulas will work. Perimeter and Area should be as follows:
P = 3(4x1² + (6 x 1) + 2) = 36
A = 9 x (2n + 1) x (2n² + 2n)
A = 9 x 4 x 3
Using the lengths of the sides added together
P = 9 + 12 + 15 = 36
A = 0.5 x 9 x 12 = 54
This shows the area formula does not work, however the perimeter formula does.
