Infinite sequences of PPT’s, generated by relatively simple formulae, can be found in a binary vine-like structure which sits ‘inside’ the ternary tree structure of PPT’s discovered by B. Berggren in 1934.

On reading an excerpt from a book ‘The Pythagorean Theorem: A 4,000-Year History (2007) by Eli Maor’ there is mention of the PPT (4601,4800,6649) which was/is stamped into the clay tablet designated Plimpton 322.

It is straightforward to calculate where this PPT sits within the ‘vine-like’ structure of all PPT’s.

(4601,4800,6649) is the first PPT (n=1) on the branch d=2(c+a) seeded by (704,903,1145) = (a,b,c).

Likewise, (704,903,1145) is the first PPT (n=1) on the branch d=2(c+a) seeded by (21,220,221) = (a,b,c).

(21,220,221) is at position n=10 on the main stem.

So, to get to (4601,4800,6649) from the root (0,1,1), take 10 steps along the main stem to (21,220,221), then take one step down the d=2(c+a) branch seeded by (21,220,221) to (704,903,1145), followed by one step down the d=2(c+a) branch seeded by (704,903,1145) to (4601,4800,6649).

Some identities

The ‘main stem’ sequence of PPT’s is just (2n+1, n(2n+1)+n, n(2n+1) +n + 1), n∈ω, where 2n+1 is any odd positive integer.

If we choose (a,b,c) = (3,4,5) as a seed PPT, then choose d=2(c+a) we get the generated sequence of PPT’s (16n+4,n(16n+8)-3,n(16n+8)+5) = (16n+4,16n²+8n-3,16n²+8n+5),n∈ω.

So, we have from the main stem, (2n+1)² = (2n² + 2n + 1)² – (2n² + 2n)² , while we also have from (16n+4,16n²+8n-3,16n²+8n+5), n∈ω that (16n²+8n-3)² = (16n²+8n+5)² – (16n+4)².

Now, 16n²+8n-3 = 2(8n²+4n-2) + 1 and so we can replace n by 8n²+4n-2 in

(2n+1)² = (2n² + 2n + 1)² -(2n² + 2n)² to get

(16n²+8n-3)² = (8(4n² +2n -1)² +4(4n² +2n -1) + 1)² -(8(4n² +2n -1)² +4(4n² +2n -1))².

Hence we have …

(16n²+8n+5)² – (16n+4)² = (8(4n²+2n-1)²+4(4n²+2n-1)+1)² – (8(4n²+2n-1)² +4(4n²+2n -1))², n∈ω.

If n=1 then this becomes 29² – 20² = 221² – 220².

If n=10 then we get 1685² – 164² = 1406165² – 1406164².

Returning to (2n² + 2n + 1)² = (2n+1)² + (2n² + 2n)² and

(16n²+8n+5)² = (16n²+8n-3)² + (16n+4)² if we write 16n²+8n+5 as 2(8n²+4n+2)+1 and substitute

8n²+4n+2 for n in (2n² + 2n + 1)² = (2n+1)² + (2n² + 2n)² we can obtain an identity to generate some Pythagorean quadruples.

This follows …

(8(4n²+2n+1)² + 4(4n²+2n+1)+1)² = (16n+4)² +(16n²+8n-3)² +(8(4n²+2n+1)² + 4(4n²+2n+1))²

Again if n = 1, we get 421² = 20² + 21² + 420².

Also if n=10, 1419613² = 164² + 1677² + 1419612².

Fermat’s PPT: this is the smallest PPT such that the ‘hypotenuse’ and the ‘sum of the other 2 sides’ are squares, i.e. (1061652293520,4565486027761,4687298610289).

A generating sequence linking this back to the main stem is easily derived …

P = 1061652293520

Q = 4565486027761

R = 4687298610289

a = 29729679503

b = 87151633296

c = 92082903025

n = 4, d=2(c+a)

P = 29729679503

Q = 87151633296

R = 92082903025

a = 7683073116

b = 10004600587

c = 12614342845

n = 2, d=2(c-a)

P = 7683073116

Q = 10004600587

R = 12614342845

a = 142061129

b = 2463588600

c = 2467681129

n = 1, d=2(c+a)

P = 142061129

Q = 2463588600

R = 2467681129

a = 1008024

b = 2915143

c = 3084505

n = 17, d=2(c+a)

P = 1008024

Q = 2915143

R = 3084505

a = 237943

b = 330576

c = 407305

n = 2, d=2(c-a)

P = 237943

Q = 330576

R = 407305

a = 8148

b = 84485

c = 84877

n = 1, d=2(c-a)

P = 8148

Q = 84485

R = 84877

a = 75

b = 308

c = 317

n = 10, d=2(c+a)

P = 75

Q = 308

R = 317

a = 3

b = 4

c = 5

n = 4, d=2(c+b)