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Pythagorean Triple Calculator & Checker

Check if three numbers form a Pythagorean triple. Identifies primitive and non-primitive triples with the base triple and scaling factor.

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Ladder Against Wall

A 10-foot ladder leans against a wall with the base 6 feet away. How high does it reach?

Key values: Leg a = 6 ft · Hypotenuse c = 10 ft · Solve for leg b

TV Screen Diagonal

Find the diagonal of a TV with a 16:9 aspect ratio (width 40", height 22.5").

Key values: Width = 40 in · Height = 22.5 in · Find diagonal

Classic 3-4-5 Triangle

Verify that sides 3, 4, and 5 form a right triangle (the most famous Pythagorean triple).

Key values: Sides: 3, 4, 5 · Right triangle · Primitive triple

Documentation

What Are Pythagorean Triples?

A Pythagorean triple is a set of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ . These are right triangles with integer side lengths—no irrational numbers, no approximations.

A triple is primitive if the three numbers share no common factor (GCD = 1). Every Pythagorean triple is either primitive or a scalar multiple of a primitive triple. For example, (6, 8, 10) = 2 × (3, 4, 5).

Euclid's Generating Formula

Every primitive Pythagorean triple can be generated using two parameters $m$ and $n$ where $m > n > 0$ :

a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2

The triple is primitive when $m$ and $n$ are coprime (share no common factor) and are of opposite parity (one even, one odd).

Why it works: The formula ensures $a^2 + b^2 = (m^2-n^2)^2 + (2mn)^2 = m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4 = (m^2+n^2)^2 = c^2$ .

Common Pythagorean Triples

Triple (a, b, c)	Primitive?	m, n	Notes
(3, 4, 5)	Yes	m=2, n=1	Smallest triple, used in construction
(5, 12, 13)	Yes	m=3, n=2	Common in textbooks
(8, 15, 17)	Yes	m=4, n=1
(7, 24, 25)	Yes	m=4, n=3
(20, 21, 29)	Yes	m=5, n=2	Near-isosceles
(6, 8, 10)	No	—	2 × (3, 4, 5)
(9, 12, 15)	No	—	3 × (3, 4, 5)

Properties of Pythagorean Triples

There are infinitely many primitive Pythagorean triples (proven by Euclid's formula: there are infinitely many valid pairs of coprime $m, n$ ).
In every primitive triple, exactly one of $a$ or $b$ is even (specifically, $b = 2mn$ is always even).
In every primitive triple, exactly one of $a$ , $b$ , $c$ is divisible by 5.
The product $abc$ is always divisible by 60 for any primitive triple.
The ancient Babylonians knew about Pythagorean triples: the Plimpton 322 tablet (c. 1800 BCE) lists 15 triples, including (4961, 6480, 8161).

Beyond Triples: Fermat's Last Theorem

While $a^2 + b^2 = c^2$ has infinitely many integer solutions, Fermat's Last Theorem (proven by Andrew Wiles in 1995) states that for any integer exponent $n > 2$ , the equation $a^n + b^n = c^n$ has no positive integer solutions. The jump from squares to cubes eliminates all solutions entirely.

Frequently Asked Questions

What is a Pythagorean triple?

A Pythagorean triple is a set of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ . These represent right triangles with integer side lengths. The smallest example is (3, 4, 5) since $9 + 16 = 25$ .

What is the difference between primitive and non-primitive triples?

A primitive triple has GCD = 1, meaning the three numbers share no common factor. A non-primitive triple is a scalar multiple of a primitive one. For example, $(6, 8, 10) = 2 \times (3, 4, 5)$ is non-primitive, while (3, 4, 5) itself is primitive.

How does Euclid's formula generate Pythagorean triples?

For integers $m > n > 0$ , the formulas $a = m^2 - n^2$ , $b = 2mn$ , $c = m^2 + n^2$ always produce a Pythagorean triple. The triple is primitive when $m$ and $n$ are coprime and of opposite parity (one even, one odd).

Are there infinitely many Pythagorean triples?

Yes. Since there are infinitely many valid pairs of coprime integers $m$ and $n$ with opposite parity, Euclid's formula generates infinitely many primitive triples. Scaling each by any positive integer produces infinitely many non-primitive triples as well.

Can the Pythagorean equation be solved with higher exponents?

No. Fermat's Last Theorem, proved by Andrew Wiles in 1995, states that $a^n + b^n = c^n$ has no positive integer solutions for any exponent $n > 2$ . The jump from squares to cubes eliminates all solutions entirely.

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