A Pythagorean triple is a set of three positive integers which satisfy the Pythagorean equation (A^{2} + B^{2} = C^{2}). To 'satisfy' an equation means that the values can be substituted into the equation and the equation remains true. The smallest of the Pythagorean triples is (3,4,5). To verify that this triple is a Pythagorean triple, substitute the numbers into the Pythagorean equation: 3^{2} + 4^{2} = 5^{2}. Now simplify the exponents: 9 + 16 = 25. Finally, simplify the addition: 25 = 25. Since the equation 25 = 25 is true, (3,4,5) qualifies as a Pythagorean triple.
A primitive Pythagorean triple is a triple where a, b, and c are coprime. Integers are coprime if they have no common factors greater than 1.
The Pythagorean Theorem was one of the earliest achievements in analytic geometry. The ancients, however, had not discovered real numbers. They had only integers and rational numbers to work with. Ancient mathematicians explored Pythagorean triples as a way of finding integer values for the lengths of the sides of right triangles.
In exploring the possibilities of Pythagorean triples, a number of formulas were invented that allowed mathematicians to quickly find many Pythagorean triples. One of these formulas is called Euclid’s formula for generating Pythagorean triples. It is: A = 2mn, B = m^{2}- n^{2}, and C = m^{2}+ n^{2}, where m and n are any positive integers such that m > n, A and B are the lengths of the legs of the right triangle, and C is the length of the hypotenuse of the right triangle.
To show that these formulas work, substitute the formulas into the Pythagorean equation:
Equation(s) | Description |
---|---|
A^{2} + B^{2} = C^{2} | Start with the Pythagorean equation. |
2mn^{2} + (m^{2}-n^{2})^{2} = (m^{2}+n^{2})^{2} | Substitute 2mn in for A, m^{2} - n^{2} in for B, and m^{2} + n^{2} in for C. |
4m^{2}n^{2} + m^{4} - 2m^{2}n^{2} +n^{4} = m^{4} + 2m^{2}n^{2} + n^{4} | Expand the exponents. |
m^{4} + 2m^{2}n^{2} + n^{4} = m^{4} + 2m^{2}n^{2} + n^{4} | Combine like terms. Since both sides of the equations are the same, this is an identity. This means that the formula will always generate Pythagorean triples. |
Table 1: Euclid's formula for generating Pythagorean triples |
While this formula will not find all Pythagorean triples. It will find infinitely many. The following table shows some of the Pythagorean triples that can be found using this formula:
m | n | A | B | C | Verification |
---|---|---|---|---|---|
2 | 1 | 2·2·1 = 4 | 2^{2} - 1^{2} = 4 - 1 = 3 | 2^{2}+ 1^{2} = 4 + 1 = 5 | 4^{2} + 3^{2} = 5^{2} → 16 + 9 = 25 → 25 = 25 |
3 | 1 | 2·3·1 = 6 | 3^{2} - 1^{2} = 9 - 1 = 8 | 3^{2}+ 1^{2} = 9 + 1 = 10 | 6^{2} + 8^{2} = 10^{2} → 36 + 64 = 100 → 100 = 100 |
3 | 2 | 2·3·2 = 12 | 3^{2}- 2^{2} = 9 - 4 = 5 | 3^{2}+ 2^{2} = 9 + 4 = 13 | 12^{2} + 5^{2} = 13^{2} → 144 + 25 = 169 → 169 = 169 |
4 | 1 | 2·4·1 = 8 | 4^{2}- 1^{2} = 16-1 = 15 | 4^{2}+ 1^{2} = 16 + 1 = 17 | 8^{2} + 15^{2} = 17^{2} → 64 + 225 = 2289 → 289 = 289 |
Table 2: Examples of Euclid's formula for generating Pythagorean triples |
When one has found a Pythagorean triple, say (a, b, c), other Pythagorean triples can be found in the form (ka, kb, kc) where k is an integer greater than 1. However, Pythagorean triples generated by this method are not primitive. This can be proved by putting ka, kb, and kc into the Pythagorean theorem:
Step | Equations | Description |
---|---|---|
1 | (ka)^{2} + (kb)^{2} = (kc)^{2} ? | This is the claim. |
2 | k^{2}a^{2} + k^{2}b^{2} = k^{2}c^{2} | Apply the distributive property of exponents. |
3 | k^{2}(a^{2} + b^{2}) = k^{2}c^{2} | Apply the distributive property of multiplication over addition and subtraction. |
4 | (k^{2}/k^{2})(a^{2}+b^{2}) = (k^{2}/k^{2})c^{2} | Use the multiplicative property of equality to multiply both sides by 1/(k^{2}). |
5 | 1(a^{2}+b^{2}) = 1(c^{2}) | Use the identity (m/m) = 1, m ≠ 0 to replace (k^{2}/k^{2}) with 1. |
6 | a^{2}+b^{2} = c^{2} | Apply the property of multiplying by 1. The equation is now identical to the Pythagorean theorem. |
Table 3: Proof of that multiples of Pythagorean triples are also Pythagorean triples. |
For example, take the Pythagorean triple (3,4,5). Some Pythagorean triples that can be derived from this triple by multiplication are (6,8,10), (9,12,15), (12,16,20), …. A Pythagorean triple that is not a multiple of another Pythagorean triple is called a primitive Pythagorean triple. The first sixteen primitive Pythagorean triples are:
( 3 , 4 , 5) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
In 1934 B. Berggren shows that all primitive Pythagorean triples can be generated from the triple (3, 4, 5). This recursive algorithm generates three Pythagorean triples for each triple it starts with. Each of those three triples then generates three more. If a and b are the legs of a Pythagorean triple and c is the hypotenuse:
new side a | new side b | new side c | |
---|---|---|---|
T1: | a - 2b + 2c | 2a - b + 2c | 2a - 2b + 3c |
T2: | a + 2b + 2c | 2a + b + 2c | 2a + 2b + 3c |
T3: | -a + 2b + 2c | -2a + b + 2c | -2a + 2b + 3c |
Table 4: Berggren's algorithm for generating Pythagorean triples. |
Start with (3, 4, 5).
new side a | new side b | new side c | |
---|---|---|---|
T1: | 3 - (2×4) + (2×5) = 5 | (2×3) - 4 + (2×5) = 12 | (2×3) - (2×4) + (3×5) = 13 |
T2: | 3 + (2×4) + (2×5) = 21 | (2×3) + 4 + (2×5) = 20 | (2×3) + (2×4) + (3×5) = 29 |
T3: | -3 + (2×4) + (2×5) = 15 | -(2×3) + 4 + (2×5) = 8 | -(2×3) + (2×4) + (3×5) = 17 |
Table 5: Example of Berggren's algorithm for generating Pythagorean triples. |
The properties of a primitive Pythagorean triple include:
# | A | B | C | D |
E | F | G | H | I |
J | K | L | M | N |
O | P | Q | R | S |
T | U | V | W | X |
Y | Z |
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