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This page shows that the Pythagorean Theorem can also be rewritten as the portions of a specific type of quadratic formula—namely, those that are the exact square of a linear function.

A Brief Introduction Of The Pythagorean Theorem

The Pythagorean Theorem, states that the squares of the measurements of the two shorter sides (the legs of a right triangle will always equal the longest side (known as the hypotenuse of the same. The theorem is usually written as &pyth;, and illustrated so:

The most commonly used numbers for this are: 3&sqrd; + 4&sqrd; = 5&sqrd;, which works out to 9 + 16 = 25, though an infinite number of such ratios—known as Pythagorean triples—exist:

If you want more proof of a infinite number existing, consider this: The difference between any square number and the next or previous square is always an odd number, and as the square of any odd number is itself an odd number, there are an infinite number of situations where that difference is itself a square number.

Linear Functions And The Squares Thereof

A Linear Function is written thusly: f(x) = ax + b. To avoid confusion later on, I’ll write use m and n in place of a and b respectively, like this: f(x) = mx + n. Squaring a linear function gives you this equation: (mx + n)&sqrd; = m&sqrd;x&sqrd; + 2mnx + n&sqrd;

If we flip the last equation around on the equals sign and simplify the portion m&sqrd;x&sqrd;, we get: (mx)&sqrd; + 2mnx + n&sqrd; = (mx + n)&sqrd;.

Comparing this to the Pythagorean Theorem, the following two equivalencies seem obvious:

This leaves b&sqrd; from the Pythagorean Theorem, and and the rest of the linear function’s square 2mnx + n&sqrd;. If b&sqrd; = 2mnx + n&sqrd;, then b = &pyb;. With this in mind, we can potentially reïllustrate the Pythagorean Theorem in relation to the triangle like this:

Since the formula &pyb; (b in the Pythagorean Theorem) is the only one involving a square root, finding an integer solution for that automatically results in integer for both a and c—and thus yeilding a Pythagorean triple.

Examples

m = 1, n = 1

Just a quick note about algebra: when a variable is multiplied by 1, the 1 is usually omitted for brevity’s sake. If the 1 is standing on its own, then it needs to be there. This first example shows what I mean.

&m2n3;

&m5n11;

An Inherent Pattern?

Pythagorean triples for &m1n1;
`x`&sqrd; + 2`x` + 1 = (`x` + 1)&sqrd;
`x`	`a`	`b`	`c`	&pyth;
4	&pya; = 1 × 4 = 4	&pyb; = (2 × 1 × 1 × 4) + 1&sqrd; = 8 + 1 = 9 = 3	&pyc; = (1 × 4) + 1 = 4 + 1 = 5	4&sqrd; + 3&sqrd; = 5&sqrd; 16 + 9 = 25
12	&pya; = 1 × 12 = 12	&pyb; = (2 × 1 × 1 × 12) + 1&sqrd; = 24 + 1 = 25 = 5	&pyc; = (1 × 12) + 1 = 12 + 1 = 13	12&sqrd; + 5&sqrd; = 13&sqrd; 144 + 25 = 169
24	&pya; = 1 × 24 = 24	&pyb; = (2 × 1 × 1 × 24) + 1&sqrd; = 48 + 1 = 49 = 7	&pyc; = (1 × 24) + 1 = 24 + 1 = 25	24&sqrd; + 7&sqrd; = 25&sqrd; 576 + 49 = 625
40	&pya; = 1 × 40 = 40	&pyb; = (2 × 1 × 1 × 40) + 1&sqrd; = 80 + 1 = 81 = 9	&pyc; = (1 × 40) + 1 = 40 + 1 = 41	40&sqrd; + 81&sqrd; = 41&sqrd; 1600 + 81 = 1681

Pythagorean triples for &m2n3;
4`x`&sqrd; + 12`x` + 9 = (2`x` + 3)&sqrd;
`x`	`a`	`b`	`c`	&pyth;
6	&pya; = 2 × 6 = 12	&pyb; = (2 × 2 × 3 × 6) + 3&sqrd; = 72 + 9 = 81 = 9	&pyc; = (2 × 6) + 3 = 12 + 3 = 15	12&sqrd; + 9&sqrd; = 15&sqrd; 144 + 81 = 225
18	&pya; = 2 × 18 = 36	&pyb; = (2 × 2 × 3 × 18) + 3&sqrd; = 216 + 9 = 225 = 15	&pyc; = (2 × 18) + 3 = 36 + 3 = 9	36&sqrd; + 15&sqrd; = 39&sqrd; 1,296 + 225 = 1,521
36	&pya; = 2 × 36 = 72	&pyb; = (2 × 2 × 3 × 36) + 3&sqrd; = 432 + 9 = 441 = 21	&pyc; = (2 × 36) + 3 = 72 + 3 = 75	72&sqrd; + 21&sqrd; = 75&sqrd; 5,184 + 441 = 5,625
60	&pya; = 2 × 60 = 120	&pyb; = (2 × 2 × 3 × 60) + 3&sqrd; = 720 + 9 = 729 = 123	&pyc; = (2 × 60) + 3 = 120 + 3 = 123	120&sqrd; + 27&sqrd; = 123&sqrd; 14,400 + 729 = 15,129

Pythagorean triples For&m5n11;
25`x`&sqrd; + 110`x` + 121 = (5`x` + 11)&sqrd;
`x`	`a`	`b`	`c`	&pyth;
88	&pya; = 5 × 88 = 440	&pyb; = (2 × 5 × 11 × 88) + 11&sqrd; = 9,680 + 121 = 9,801 = 99	&pyc; = (5 × 88) + 11 = 440 + 11 = 451	440&sqrd; + 99&sqrd; = 451&sqrd; 193,600 + 9,801 = 203,401
132	&pya; = 5 × 132 = 660	&pyb; = (2 × 5 × 11 × 132) + 11&sqrd; = 14,520 + 121 = 14,641 = 121	&pyc; = (5 × 132) + 11 = 660 + 11 = 672	660&sqrd; + 121&sqrd; = 671&sqrd; 435,600 + 14,641 = 450,241
396	&pya; = 5 × 396 = 1,980	&pyb; = (2 × 5 × 11 × 396) + 11&sqrd; = 43,560 + 121 = 43,681 = 209	&pyc; = (5 × 396) + 11 = 1,980 + 11 = 1,991	1,980&sqrd; + 209&sqrd; = 1,991&sqrd; 3,920,400 + 43,681 = 3,964,081
484	&pya; = 5 × 484 = 2,420	&pyb; = (2 × 5 × 11 × 484) + 11&sqrd; = 53,240 + 121 = 53,361 = 231	&pyc; = (5 × 484) + 11 = 2,420 + 11 = 2,431	2,420&sqrd; + 231&sqrd; = 2,431&sqrd; 5,856,400 + 53,361 = 5,909,761

When I first started looking at other solutions to the Pythagorean Theorem, I discovered an apparent pattern: if the numbers were co-prime, then c = a + 1. Co-prime means that a set of numbers do not share prime factors between them. A good example is the Pythagorean triple 21, 220, and 221. None of these are prime numbers:

However, no prime factor is shared between them (these are known as primitive Pythagorean triples). This is not the case for &m2n3; and &m5n11;:as shown blow, all the numbers in the Pythagorean triples for those two equations share the prime factors 3 and 11 respectively:

&m2n3;
Original Equation	Dividing `a`, `b`, and `c` by 3
12&sqrd; + 9&sqrd; = 15&sqrd;	4&sqrd; + 3&sqrd; = 5&sqrd;
36&sqrd; + 15&sqrd; = 39&sqrd;	12&sqrd; + 5&sqrd; = 13&sqrd;
72&sqrd; + 21&sqrd; = 75&sqrd;	24&sqrd; + 7&sqrd; = 25&sqrd;
120&sqrd; + 27&sqrd; = 123&sqrd;	40&sqrd; + 9&sqrd; = 41&sqrd;

&m5n11;
Original Equation	Dividing `a`, `b`, and `c` by 11
440&sqrd; + 99&sqrd; = 451&sqrd;	40&sqrd; + 9&sqrd; = 41&sqrd;
660&sqrd; + 121&sqrd; = 671&sqrd;	60&sqrd; + 11&sqrd; = 61&sqrd;
1,980&sqrd; + 209&sqrd; = 1,991&sqrd;	180&sqrd; + 19&sqrd; = 181&sqrd;
2,420&sqrd; + 231&sqrd; = 2,431&sqrd;	220&sqrd; + 21&sqrd; = 221&sqrd;

In other words, those triples are just multiples of &pytha1;, and it looked like like the real way to write the Pythagorean Theorem was (ax)&sqrd; + (bx)&sqrd; = (x(a + 1))&sqrd;.

Then I tried &m1n2; and I found some primitive Pythagorean triples in the pattern a&sqrd; + b&sqrd; = (a + 2)&sqrd;.

&m1n2;
`x`&sqrd; + 4`x` + 4 = (`x` + 2)&sqrd;
`x`	`a`	`b`	`c`	&pyth;
15	&pya; = 1 × 15 = 15	&pyb; = (2 × 1 × 2 × 15) + 2&sqrd; = 60 + 4 = 64 = 8	&pyc; = (1 × 15) + 2 = 15 + 2 = 17	15&sqrd; + 8&sqrd; = 17&sqrd; 225 + 64 = 289
35	&pya; = 1 × 35 = 35	&pyb; = (2 × 1 × 2 × 35) + 2&sqrd; = 140 + 4 = 144 = 12	&pyc; = (1 × 35) + &sqrd; = 35 + 2 = 37	35&sqrd; + 12&sqrd; = 37&sqrd; 1,225 + 144 = 1,369
63	&pya; = 1 × 63 = 63	&pyb; = (2 × 1 × &sqrd; × 63) + 2&sqrd; = 252 + 4 = 256 = 16	&pyc; = (1 × 63) + 2 = 63 + 2 = 65	63&sqrd; + 16&sqrd; = 65&sqrd; 3,969 + 256 = 4,225
99	&pya; = 1 × 99 = 99	&pyb; = (2 × 1 × 2 × 99) + 2&sqrd; = 396 + 4 = 400 = 20	&pyc; = (1 × 99) + 2 = 99 + 2 = 101	99&sqrd; + 20&sqrd; = 101&sqrd; 9,801 + 400 = 10,201

Other Fun Facts About Primitive Pythagorean Triples

Quick note: a and b refer to the legs of a Pythagorean Triangle, while c is the hypotenuse.

(c-a)(c-b)2 is always a perfect square.
c is always an odd number in the form of 4n+1 (in other words, c - 1 is always a multiple of 4). If c is a composite number, all its prime factors can be expressed the same way.
One of either a or b is a multiple of 3, but never c.
Whichever of a or b is even is also a multiple of 4.
Exactly one of a, b or c is a multiple of 5.
it is possible for a or b to be
- A multiple of 12 (3×4)
- A multiple of 15 (3×5) (which means the other will be a multiple of 4)
- A multiple of 20 (4×5) (which means the other will be a multiple of 3)
- A multiple of 60 (3×4×5)
Speaking of 60, that’s the largest number that always devides into abc
At most one of a, b or c is a square number (for example, in the triple (7, 24, 25), 25 = 5&sqrd;). Usually, none of them are.
Since one of the numbers is always divisible by 4 (a composite number), at most only 2 of a, b or c can be prime. In some triples, such as (21,220,221) and (27, 364, 365), none of them are prime.
While the hypotenuse from a Pythagorean triple will invariably pop up as one of the legs of another such triangle (for example: (3,4,5), (5,12,13), (13,84,85), etc.), there are no Pythagorean triangles where one of the legs and the hypotenuse show up as the legs of another Pythagorean triangle.
Every integer > 2 not in the form of 4n+2 (that is, divisible by 2, but not by 4) is part of at least one primitive Pythagorean triple. If non-primitive Pythagorean triples (triples in which all three numbers share at least one prime factor) are included, then every integer > 2 is part of at least one Pythagorean triple.