Pythagorean Triplets: Pythagoras, the famous Greek philosopher, gave a beautiful relation between the lengths of sides of a right-angled triangle which is generally known as Pythagoras theorem, which states that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of its remaining two sides.

A right-angled triangle is a triangle whose one angle is a right angle, i.e., which measures \({\rm{90}}^\circ .\)

Three positive numbers \(x,y\) and \(z\) are said to be a Pythagorean triplet, if \({z^2} = {x^2} + {y^2}.\) This article will discuss the Pythagorean triplets and the numbers that form the Pythagorean triplets.

Right Angled Triangle

A triangle, one of whose angles measures \({\rm{90}}^\circ ,\) is called a right-angled triangle.
The right-angled triangle’s side opposite to the right angle is known as the triangle’s hypotenuse. The other two sides are called the base or adjacent and opposite or altitude or perpendicular of the right-angled triangle.

Properties of Right Angled Triangle

1. Any one of the angles of a right-angled triangle always measures \({\rm{90}}^\circ .\)
2. The opposite side of the angle \({\rm{90}}^\circ \) is called the hypotenuse.
3. The hypotenuse is the longest side of the right-angled triangle.
4. The sum of the other two interior angles of a right-angled triangle always measures \({\rm{90}}^\circ .\)

Pythagoras Theorem

Pythagoras was an eminent Greek philosopher born in \(580{\rm{\;BC\;}}\) and died in \(500{\rm{\;BC\;}}\) who has given a wonderful relation between the lengths of the sides of a right-angled triangle, known as Pythagoras theorem.

Statement: In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of its remaining two sides.

Thus, if \(\Delta ABC\) is a right-angled triangle at \(C,\) so that \(AB\) is the hypotenuse and \(AC\) and \(BC\) are the sides of the right angle, then
\({\left( {AB} \right)^2} = {\left( {BC} \right)^2} + {\left( {CA} \right)^2}\)
i.e. \({\left( {{\rm{Hypotenuse}}} \right)^2} = {\left( {{\rm{Base}}} \right)^2} + {\left( {{\rm{perpendicular}}} \right)^2}\)

To prove the above theorem, we use another theorem, which states, “If a perpendicular is drawn from the vertex of the right-angled triangle to the hypotenuse, then triangles on both sides of the perpendiculars are similar to the whole triangle and to each other.”

Let us now apply this theorem in proving the Pythagoras theorem.

Pythagoras Theorem Proof

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
We are given a right triangle \(ABC\) right-angled at \(B.\)
We need to prove that \(A{C^2} = A{B^2} + B{C^2}\)
Let us draw \(BD \bot AC\)

Now, \(\Delta ADB \sim \Delta ABC\) (from above theorem)
So, \(\frac{{AD}}{{AB}} = \frac{{AB}}{{AC}}\) (sides are proportional)
\(\Rightarrow AD \times AC = A{B^2}\,\,\,\,\,…\left( 1 \right)\)
Also, \(\Delta BDC \sim \Delta ABC\) (from above theorem)
So, \(\frac{{CD}}{{BC}} = \frac{{BC}}{{AC}}\)
\(\Rightarrow CD \times AC = B{C^2}\,\,\,…\left( 2 \right)\)
Adding \(\left( 1 \right)\) and \(\left( 2 \right),\) we get
\(\left( {AD \times AC} \right) + \left( {CD \times AC} \right) = A{B^2} + B{C^2}\)
\(\Rightarrow AC\left( {AD + CD} \right) = A{B^2} + B{C^2}\)
\(\Rightarrow AC \times AC = A{B^2} + B{C^2}\)
\(\Rightarrow A{C^2} = A{B^2} + B{C^2}\)

The above theorem was earlier given by an ancient Indian Mathematician Baudhayan about \(800\,{\rm{BC}}\) in the following form.

“The diagonal of a rectangle produces by itself the same area as produced by its both sides, i.e., length and breadth.”

For this reason, the Pythagoras theorem is sometimes also referred to as the Baudhayan Theorem.

Pythagorean Triplets

Three positive numbers \(a,\;b,\;c\) in this order, are said to form a Pythagorean triplet, if \({c^2} = {a^2} + {b^2}\) Here \(a\) is the perpendicular, \(b\) is the base, and \(c\) is the hypotenuse of the right-angled triangle.

Pythagorean triplets refer to a set of \(3\) numbers or integers which satisfy the rule of Pythagoras theorem. Pythagoras theorem relates the squares on the sides of a right-angled triangle. Pythagorean triplets will have a set of three positive integers such that the square of the largest among the three integers is equal to the sum of the squares of the other two integers.

Types of Pythagorean Triplets

There are two types of Pythagorean triplet, they are:

1. Primitive Pythagorean triplet
2. Non-primitive Pythagorean triplet

Primitive Pythagorean Triplet

A primitive Pythagorean triplet is such that all the three numbers or integers in the triplets do not have any common divisor other than \(1\) or any one of the triplets can have only one even positive integer among three. In other words, consider a Pythagorean triplet \(\left( {a,\;b,c} \right)\) such that HCF of \(\left( {a,\;b,\;c} \right) = 1\)
Example: \(\left( {3,\;4,\;5} \right)\) is a primitive Pythagorean triplet

Non-primitive Pythagorean triplet

A non-primitive Pythagorean triplet follows the Pythagoras theorem and also has a common divisor. Non-primitive Pythagorean triplet may or may not have more than one even positive integer.
Example: \(6,\;8,\;10\)

Generating a Pythagorean Triplet

Euclid’s formula is the fundamental formula for generating the Pythagorean triplets given an arbitrary pair of integers \(m\) and \(n\) with \(m > n > 0.\) The formula states that the integers \(a = {m^2} – {n^2},b = 2\,{\rm{mn}},c = {m^2} + {n^2}\) form a Pythagorean triplet. The triplet generated by Euclid’s formula is primitive if and only if \(m\) and \(n\) are coprime and not both odd. When both \(m\) and \(n\) are odd, then \(a,\;b,\) and \(c\) will be even, and the triplet will not be primitive. However, dividing \(a,\;b,\) and \(c\) by \(2\) will give a primitive triplet when \(m\) and \(n\) are coprime and odd.

Every primitive triplet appears (after the exchange of \(a\) and \(b,\) if \(a\) is even) from a unique pair of coprime numbers \(m,\;n\) one of which is even. It follows that there are infinitely many primitive Pythagorean triplets.

Example: Let us try generating a Pythagorean triplet using the two integers \(3\) and \(5.\) Since \(m > n,\,m\; = \;5,\) and \(n\; = \;3.\) Since we know the values of \(m\) and \(n,\) substituting those values into the formulas of \(a,b\) and \(c,\) to get the sides of the right triangle.
Calculating \(a\) : \(a = {m^2} – {n^2}\)
\(a = {5^2} – {3^2}\)
\(a = 25 – 9\)
\(a = 16\)
Calculating \(b\) : \(b = 2\,{\rm{mn}}\)
\(\Rightarrow b = 2 \times 5 \times 3\)
\(\Rightarrow b = 30\)
Calculating \(c\): \(c = {m^2} + {n^2}\)
\(\Rightarrow c = {5^2} + {3^2}\)
\(\Rightarrow c = 25 + 9\)
\(\Rightarrow c = 34\)
Substituting the \(a,b\) and \(c\) value in the general form of Pythagorean triplet \({c^2} = {a^2} + {b^2},\) we get
\({\left( {34} \right)^2} = {\left( {16} \right)^2} + {\left( {30} \right)^2}\)
\(\Rightarrow 1156 = 256 + 900\)
\(\Rightarrow 1156 = 1156\)
Therefore, \(\left( {16,\;30,\;34} \right)\) are Pythagorean triplets.

Table of Pythagorean Triplets

There are \(16\) primitive Pythagorean triplets of numbers upto \(100\)

\(3,\;4,\;5\)	\(5,\;12,\;13\)	\(8,\;15,\;17\)	\(7,\;24,\;25\)
\(20,\;21,\;29\)	\(12,\;35,\;37\)	\(9,\;40,\;41\)	\(28,\;45,\;53\)
\(11,\;60,\;61\)	\(16,\;63,\;65\)	\(33,\;56,\;65\)	\(48,\;55,\;73\)
\(13,\;84,\;85\)	\(36,\;77,\;85\)	\(39,\;80,\;89\)	\(65,\;72,\;97\)

Other small Pythagorean triplets such as \(6,\;8,\;10\) are not listed because they are non-primitive, and they are multiple of \(3,\;4,\;5.\)

The list of Pythagorean triplets where the value of hypotenuse is more than \(100\) is given below:

\(20,\;99,\;101\)	\(60,\;91,\;109\)	\(15,\;112,\;113\)	\(44,\;117,\;125\)
\(88,{\rm{\;}}105,{\rm{\;}}137\)	\(17,{\rm{\;}}144,{\rm{\;}}145\)	\(24,{\rm{\;}}143,{\rm{\;}}145\)	\(51,{\rm{\;}}140,{\rm{\;}}149\)
\(85,{\rm{\;}}132,{\rm{\;}}157\)	\(119,{\rm{\;}}120,{\rm{\;}}169\)	\(52,{\rm{\;}}165,{\rm{\;}}173\)	\(19,{\rm{\;}}180,{\rm{\;}}181\)
\(57,{\rm{\;}}176,{\rm{\;}}185\)	\(104,{\rm{\;}}153,{\rm{\;}}185\)	\(95,{\rm{\;}}168,{\rm{\;}}193\)	\(28,{\rm{\;}}195,{\rm{\;}}197\)
\(84,{\rm{\;}}187,{\rm{\;}}205\)	\(133,{\rm{\;}}156,{\rm{\;}}205\)	\(21,{\rm{\;}}220,{\rm{\;}}221\)	\(140,{\rm{\;}}171,{\rm{\;}}221\)
\(60,{\rm{\;}}221,{\rm{\;}}229\)	\(105,{\rm{\;}}208,{\rm{\;}}233\)	\(120,{\rm{\;}}209,{\rm{\;}}241\)	\(32,{\rm{\;}}255,{\rm{\;}}257\)
\(23,{\rm{\;}}264,{\rm{\;}}265\)	\(96,{\rm{\;}}247,{\rm{\;}}265\)	\(69,{\rm{\;}}260,{\rm{\;}}269\)	\(115,{\rm{\;}}252,{\rm{\;}}277\)
\(160,{\rm{\;}}231,{\rm{\;}}281\)	\(161,{\rm{\;}}240,{\rm{\;}}289\)	\(68,{\rm{\;}}285,{\rm{\;}}293\)

Solved Examples – Pythagorean Triplets

Q.1. Prove that \(\left( {9,\;40,\;41} \right)\) is a Pythagorean triplet.
Ans: Let \(\left( {a,b,c} \right)\) be the Pythagorean triplet. Then, \({c^2} = {a^2} + {b^2}\)
Let \(a = 9,\;b = 40\) and \(c = 41\)
So, \({\left( {41} \right)^2} = {\left( 9 \right)^2} + {\left( {40} \right)^2}\)
\(\Rightarrow 1681 = 81 + 1600\)
\(\Rightarrow 1681 = 1681\)
So, LHS = RHS
Therefore, \(\left( {9,\;40,\;41} \right)\) is a Pythagorean triplet.

Q.2. Verify whether \(\left( {4,\;5,\;16} \right)\) are Pythagorean triplets.
Ans: Let \(\left( {a,b,c} \right)\) be the Pythagorean triplet. Then, \({c^2} = {a^2} + {b^2}\)
Let, \(a = 4,\;b = 5\) and \(c = 16\)
So, \({16^2} = {4^2} + {5^2}\)
\(\Rightarrow 256 = 16 + 25\)
\(\Rightarrow 256 \ne 41\)
Hence, the given set of integers does not satisfy the Pythagoras theorem. So, \(\left( {4,\;5,\;16} \right)\) is not a Pythagorean triplet.

Q.3. Shivu tried a new route to reach his school today. He walked \(10\) blocks to the north and then \(24\) blocks to the west. Can you find how far his school is from his home?
Ans: The distance from the school to home is the length of the hypotenuse. Let \(c\) be the missing distance from school to home and \(a\; = \;10,\,b\; = \;24.\)
We know that, \({c^2} = {a^2} + {b^2}\)
\(\Rightarrow {c^2} = {10^2} + {24^2}\)
\(\Rightarrow {c^2} = 100 + 576\)
\(\Rightarrow {c^2} = 676\)
\(\Rightarrow c = \sqrt {676}\)
\(\Rightarrow c = 26\)
Therefore, the distance from school to home is \(26\) blocks.

Q.4. John’s mother asked him to place the ladder at such a distance from the wall, such that the \(17\) feet tall ladder’s head falls precisely on the top of the \(15\) feet wall. Can you help John to find the distance of the ladder from the wall?
Ans: Let the required distance be \(x\) feet. Here, the ladder, the wall, and the ground form a right-angled triangle. The ladder is the hypotenuse of the triangle.
We know that, \({c^2} = {a^2} + {b^2}.\) Let \(a = 15,\;b = x\) and \(c = 17\)
So, \({17^2} = {15^2} + {x^2}\)
\(\Rightarrow {x^2} = 289 – 225\)
\(\Rightarrow {x^2} = 64\)
\(\Rightarrow x = \sqrt {64}\)
\(\Rightarrow x = 8\)
Therefore, the distance of the ladder from the wall is \(8\) feet.

Q.5. : Using \(2\) and \(3\) as the integers, generate a Pythagorean triplet.
Ans: Since \(m > n,\) let \(m\; = \;3,\) and \(n\; = \;2.\) Since we know the values of \(m\) and \(n,\) substituting those values into the formulas of \(a,\;b,\) and \(c,\) to get the sides of the right triangle.
Calculating \(a\) : \(a = {m^2} – {n^2}\)
\(a = {3^2} – {2^2}\)
\(a = 9 – 4\)
\(a = 5\)
Calculating \(b\) : \(b = 2\,{\rm{mn}}\)
\(\Rightarrow b = 2 \times 3 \times 2\)
\(\Rightarrow b = 12\)
Calculating \(c\) : \(c = {m^2} + {n^2}\)
\(\Rightarrow c = {3^2} + {2^2}\)
\(\Rightarrow c = 9 + 4\)
\(\Rightarrow c = 13\)
Substituting the \(a,b\) and \(c\) value in the general form of Pythagorean triplet \({c^2} = {a^2} + {b^2},\) we get
\({\left( {13} \right)^2} = {\left( 5 \right)^2} + {\left( {12} \right)^2}\)
\(\Rightarrow 169 = 25 + 144\)
\(\Rightarrow 169 = 169\)
Therefore, \(\left( {5,\;12,\;13} \right)\) are Pythagorean triplets.

Summary

The three positive numbers that entirely satisfy the Pythagorean theorem are known as Pythagorean triples. The theorem claims that the square of the hypotenuse in every right triangle is equal to the sum of the squares of the other two legs. Pythagorean triples are formed from the three sides of a right triangle.

1. Pythagorean triples are any three positive numbers that meet the formula a² + b² = c².
2. If one of the numbers in a Pythagorean triple is known, the other two may be calculated using the formulas a = m² – n², b = 2mn, and c = m² + n².
3. Pythagorean triples can’t have decimals in them.
4. If (a, b, c) is a Pythagorean triplet, then ta, tb, tc, where t is any positive integer, will likewise be a Pythagorean triplet.

In the above article, we learned about the Pythagoras theorem, Pythagorean triplets, examples of Pythagorean triples, and generation of Pythagorean triplets and solved some examples regarding the same.

Frequently Asked Questions (FAQ) – Pythagorean Triplets

Q.1. What are the five most common Pythagorean triples?
Ans: The five most commonly used Pythagorean triplets are \(\left( {3,\;4,\;5} \right),\;\left( {5,\;12,\;13} \right),\;\left( {8,\;15,\;17} \right),\;\left( {7,\;24,\;25} \right)\) and \(\left( {20,\;21,\;29} \right)\)

Q.2. How to find Pythagorean triplets?
Ans: Euclid’s formula is the fundamental formula for generating the Pythagorean triplets given an arbitrary pair of integers \(m\) and \(n\) with \(m > n > 0.\) The formula states that the integers \(a = {m^2} – {n^2},b = 2\,{\rm{mn}},c = {m^2} + {n^2}\) form a Pythagorean triplet.

Q.3. Define Pythagorean triplet.
Ans: Three positive numbers \(a,\;b,\;c\) in this order, are said to form a Pythagorean triplet, if \({c^2} = {a^2} + {b^2}.\) Here \(a\) is the perpendicular, \(b\) is the base, and \(c\) is the hypotenuse of the right-angled triangle.

Q.4. Is \(\left( {6,\;8,\;10} \right)\) a Pythagorean triplet?
Ans: Let \(a = 6,\;b = 8\) and \(c = 10.\) We know that, \({c^2} = {a^2} + {b^2}\)
So, \({10^2} = {6^2} + {8^2}\)
\(\Rightarrow 100 = 36 + 64\)
\(\Rightarrow 100 = 100\)
Therefore, \(\left( {6,\;8,\;10} \right)\) is a Pythagorean triplet.

Q.5. Is \(\left( {4,5,8} \right)\) is Pythagorean triplet?
Ans: Let \(a = 4,\;b = 5\) and \(c = 8.\) We know that, \({c^2} = {a^2} + {b^2}\)
So, \({8^2} = {4^2} + {5^2}\)
\(\Rightarrow {4^2} + {5^2} = 16 + 25 = 41\)
\(\Rightarrow 64 \ne 41\)
Therefore, \(\left( {4,\;5,\;8} \right)\) is not a Pythagorean triplet.

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