Generating Pythagorean Triples Advanced. This lesson goes beyond core Geometry. You can skip it.

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Learning goals

Reduce a Pythagorean triple to its primitive triple and identify the scale factor
Derive and use the formulas that generate Pythagorean triples
Explain how parity and coprimality determine whether the formulas give a primitive triple
Recover the two parameters of a primitive triple and justify why the recovery works

Reduce a triple to its smallest integer copy

A Pythagorean triple consists of positive integers $a$ , $b$ , and $c$ with

a^2+b^2=c^2.

By the Pythagorean theorem and its converse, these are the lengths of a right triangle, with hypotenuse $c$ . To see that the lengths make a triangle, build a right corner with legs $a$ and $b$ . The third side has positive length $c$ by the theorem. The two leg entries can go in either order.

A triple is primitive when the greatest common factor (GCF) of its three entries is $1$ . The triangles below have the same shape. The small one has no common factor greater than $1$ , and each length of the larger one is three times its partner.

One drawing scale for both triangles shows what multiplying each side by three does to the shape.

Multiplying all lengths by a positive integer $k$ multiplies both sides of the squared equation by $k^2$ . Dividing all lengths by their GCF reverses that operation and leaves a primitive triple. The GCF must be removed in full. Any common factor still left in the quotient would give a larger common factor of the original lengths.

Worked example 1 Find the original shape

Problem. A right triangle has side lengths $36$ , $105$ , and $111$ units. Find its primitive triple and the integer scale factor from the primitive triangle to this triangle.

Solution. The common factors of $36$ and $105$ are $1$ and $3$ . Since $111$ is also divisible by $3$ , the GCF of all three lengths is $3$ .

Divide all three entries by $3$ .

\left(\frac{36}{3},\frac{105}{3},\frac{111}{3}\right) =(12,35,37).

The primitive triple is $(12,35,37)$ . The scale factor to the given triangle is $3$ .

Check your understanding

The right triangle with sides $170$ , $264$ , and $314$ has which primitive triple?

(85,132,157)

(3,4,5)

(170,264,314)

(85,132,314)

Build a triple from two squares

A sum and difference can give two sides whose squares differ by a perfect square. Start with $49$ and $4$ . Their sum is $53$ and their difference is $45$ .

Factor the difference of the new squares. In the sum, the $4$ terms cancel. In the difference, the $49$ terms cancel.

\begin{aligned} 53^2-45^2&=(53+45)(53-45)\\ &=(2\cdot49)(2\cdot4)\\ &=(2\cdot7\cdot2)^2\\ &=28^2. \end{aligned}

So $45$ , $28$ , and $53$ are side lengths of a right triangle. The product became a square because both starting numbers were squares.

Why the two-parameter construction gives a right triangle#

Take integers $m>n>0$ . Write $a$ for the difference of their squares and $c$ for the sum.

\begin{aligned} a&=m^2-n^2\\ c&=m^2+n^2 \end{aligned}

Factoring uses the same sums and differences as with $49$ and $4$ .

\begin{aligned} c^2-a^2&=(c+a)(c-a)\\ &=(2m^2)(2n^2)\\ &=(2mn)^2. \end{aligned}

Use $2mn$ as the remaining leg $b$ . The factored equation then gives $a^2+b^2=c^2$ . All entries are positive because $m>n>0$ . The choice of $7$ and $2$ in the numerical argument only supplied the two squares.

The three entries $(m^2-n^2,\ 2mn,\ m^2+n^2)$ are called Euclid’s formulas. The second entry is always even, but it may be the shorter or the longer leg.

Worked example 2 The even leg comes second

Problem. Generate the triple for the parameter pair $(8,3)$ . Give the two legs in increasing order and name the hypotenuse.

Solution. The difference gives one leg, the doubled product gives the other, and the sum gives the hypotenuse.

\begin{aligned} a&=64-9\\ a&=55\\[2pt] b&=2\cdot8\cdot3\\ b&=48\\[2pt] c&=64+9\\ c&=73. \end{aligned}

The legs in increasing order are $48$ and $55$ . The hypotenuse is $73$ .

Check your understanding

For the parameter pair $(10,3)$ , which entry equals the hypotenuse?

91

60

109

130

Tell which parameters give a primitive triple

Some pairs build a triangle whose sides have a common factor. A pair of integers is coprime if its GCF is $1$ . It has opposite parity if one integer is even and the other is odd.

Worked example 3 Two tests for the inputs

Problem. For each pair below, decide whether the output of Euclid’s formulas is primitive. When it is scaled, find the primitive triple and the scale factor.

Case A uses the parameter pair $(5,4)$ .

Case B uses the parameter pair $(5,3)$ .

Case C uses the parameter pair $(6,3)$ .

Solution. Case A gives $(9,40,41)$ . The GCF of $9$ and $40$ is $1$ , so the triple is primitive. The parameters are coprime and have opposite parity.

Case B gives $(16,30,34)$ . Both parameters are odd, so both the difference and sum of their squares are even. The GCF of all three entries is $2$ , giving primitive triple $(8,15,17)$ and scale factor $2$ .

Case C gives $(27,36,45)$ . Both parameters are multiples of $3$ , so the formulas use multiples of $3^2$ . The GCF of the three entries is $9$ , giving primitive triple $(3,4,5)$ and scale factor $9$ .

All three right triangles use the same drawing scale. An even output in every entry reveals the parity case, while a shared factor of three in the parameters supplies nine in every entry.

The formulas give a primitive triple exactly when the parameters are coprime and have opposite parity.

Why coprimality and opposite parity give a primitive triple#

If the parameters share a factor $d>1$ , each formula contains a factor $d^2$ , just as $3^2$ divided all the outputs in Case C.

If the parameters have the same parity, their square sum and square difference are even. The doubled product is already even, so $2$ divides all three outputs.

Now take coprime parameters of opposite parity. For example, with the parameter pair $(9,2)$ , both $77$ and $85$ are odd, so $2$ cannot divide the whole triple. Any odd prime dividing both $77$ and $85$ would divide their sum and difference.

\begin{aligned} 85+77&=2\cdot81\\ 85-77&=2\cdot4. \end{aligned}

That prime would have to divide both $9$ and $2$ , since it is odd and every prime factor of a square comes from its base. The two parameters are coprime.

For any coprime parameters of opposite parity, $a$ and $c$ are odd, and the same sum and difference are $2m^2$ and $2n^2$ . No odd prime can divide both because that prime would divide both $m$ and $n$ . So no prime divides all three outputs, and their GCF is $1$ .

Check your understanding

Why can the parameter pair $(12,9)$ produce a scaled triple even though they have opposite parity?

A Their squares are both odd B Every doubled product forces all three outputs to be even C A leg is longer than the hypotenuse D The shared factor 3 puts a factor 9 in each output

Recover the parameters of a primitive triple

The formulas can be reversed. For the primitive triple with legs $33$ and $56$ and hypotenuse $65$ , use the odd leg for $a$ and the even leg for $b$ . The squares we want come from adding and subtracting the odd leg.

\begin{aligned} \frac{65+33}{2}&=49\\ \frac{65-33}{2}&=16. \end{aligned}

The two results are positive and coprime. Their product is $(56/2)^2$ , a square. Since a prime can appear in at most one of the two factors, each factor must contain its own primes in pairs. Here $49$ contains two copies of $7$ , and $16$ contains four copies of $2$ . Taking positive square roots gives $7$ and $4$ , which are coprime and have opposite parity.

Adding the odd leg to the hypotenuse gives twice the larger square. Subtracting the same leg gives twice the smaller one.

Why every primitive triple has these parameters#

A primitive triple has exactly one even leg. Two even legs would make $c^2$ even and hence $c$ even, giving a common factor of $2$ . Two odd legs would make $c^2$ leave remainder $2$ on division by $4$ . A square has remainder $0$ or $1$ : an even square is divisible by $4$ , and an odd square has the form $4t(t+1)+1$ for some integer $t$ . With one odd and one even leg, $c^2$ is odd, so $c$ is odd.

Name the odd leg $a$ and the even leg $b$ . Define the positive integers

\begin{aligned} u&=\frac{c+a}{2}\\ v&=\frac{c-a}{2}. \end{aligned}

Both are integers since $a$ and $c$ are odd, and $u>v>0$ since $c>a>0$ . Factoring the difference of squares gives

\begin{aligned} uv&=\frac{(c+a)(c-a)}{4}\\ &=\frac{c^2-a^2}{4}\\ &=\left(\frac b2\right)^2. \end{aligned}

Any prime shared by $u$ and $v$ would divide their sum $c$ and their difference $a$ . It would then divide $b^2$ , since $b^2=c^2-a^2$ , and so divide $b$ as well. Primitivity excludes any such prime.

Because $uv$ is a square, every prime in its factorization occurs an even number of times. Since $u$ and $v$ share no primes, those even counts already occur in each factor separately. The factors are the squares of positive integers $m$ and $n$ , respectively, with $m>n$ . A shared prime of $m$ and $n$ would also be shared by $u$ and $v$ , so $m$ and $n$ are coprime. Their squares add to the odd integer $c$ , which means $m$ and $n$ have opposite parity.

Add and subtract the two squares to recover $c$ and $a$ . Take the positive square root of $uv$ to recover half of $b$ .

\begin{aligned} c&=m^2+n^2\\ a&=m^2-n^2\\ b&=2mn. \end{aligned}

The argument used the same sum, difference and prime-pairing steps as the numerical example. The odd leg and hypotenuse determine $u$ and $v$ , so they also determine the unique positive parameters with $m>n$ .

Worked example 4 Recover a pair from an even leg listed first

Problem. The primitive triple is $(60,11,61)$ . Find the positive integers $m>n$ in Euclid’s formulas and use them to check the even leg.

Solution. Use $11$ as the odd leg and $61$ as the hypotenuse.

\begin{aligned} m^2&=\frac{61+11}{2}\\ &=36\\[2pt] n^2&=\frac{61-11}{2}\\ &=25. \end{aligned}

Taking positive roots gives the parameter pair $(6,5)$ . The even leg checks because

2\cdot6\cdot5=60.

Check your understanding

For the primitive triple with legs $20$ and $99$ and hypotenuse $101$ , which pair gives the parameters $(m,n)$ ?

(100,1)

(10,1)

(11,9)

(10,9)

Separate the scale factor from the parameters

For any Pythagorean triple, divide by the GCF first. The quotient is primitive, so it has a unique pair of coprime, opposite-parity parameters. Then multiply the three generated entries by the original GCF $k$ to recover the given triple. Leg order can be swapped.

Worked example 5 Recover both the scale and the shape

Problem. A right triangle has sides $132$ , $351$ , and $375$ units. Find its integer scale factor from the primitive triangle, and find the parameters of that primitive triple.

Solution. The GCF of $132$ and $351$ is $3$ . Since $375$ is divisible by $3$ , the GCF of all three is $3$ . Dividing gives primitive triple $(44,117,125)$ .

Use the odd leg $117$ and the hypotenuse $125$ .

\begin{aligned} m^2&=\frac{125+117}{2}\\ &=121\\[2pt] n^2&=\frac{125-117}{2}\\ &=4. \end{aligned}

The parameters are $11$ and $2$ . Their doubled product is $44$ , which matches the even leg of the primitive triple. The original triangle uses scale factor $3$ .

Check your understanding

A right triangle has sides $50$ , $624$ , and $626$ . Why should you divide by the GCF before recovering primitive parameters?

A The scaled hypotenuse is always prime B The even leg must become the shorter side C The reverse proof needs the primitive triple's parity and absence of common factors D A scale factor leaves every length odd

Generating Pythagorean Triples Advanced. This lesson goes beyond core Geometry. You can skip it.

Learning goals

Reduce a triple to its smallest integer copy

Build a triple from two squares

Tell which parameters give a primitive triple

Recover the parameters of a primitive triple

Separate the scale factor from the parameters

Common mistakes

Practice

Multiple Choice Questions (MCQ)

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