Baudhayana (Pythagorean) triples

Some right triangles are unusually well-behaved: all three of their sides come out as whole numbers, with no square roots or decimals anywhere. The Egyptians used a $3,4,5$ loop of rope to build a perfect right angle at the corner of a temple; the Baudhāyana Śulba-sūtra in India listed the very same fact centuries before it carried the name "Pythagoras" in Greece. A Baudhāyana (Pythagorean) triple is any set of three positive whole numbers $(a,b,c)$ satisfying $a^2+b^2=c^2$ , so that they can literally be the three sides of a right-angled triangle. This article collects the standard triples every student should recognise on sight, explains what makes a triple primitive, shows how scaling multiplies triples into infinite families, and proves — with plain algebra — a formula that manufactures brand-new primitive triples from any two whole numbers you choose.

The core insight

A Pythagorean triple $(a,b,c)$ satisfies $a^2+b^2=c^2$ ; a handful of primitive triples such as $(3,4,5)$ generate infinitely many more by scaling, and the formula $(m^2-n^2,\,2mn,\,m^2+n^2)$ generates fresh primitive ones directly.

The core idea

A Baudhāyana (Pythagorean) triple is an ordered set of positive integers $(a,b,c)$ that fits the relation

a^2+b^2=c^2

where: $c$ is always the largest of the three numbers (the hypotenuse), and $a,b$ are the two shorter sides (the legs). If three whole numbers satisfy this equation, they can serve as the side lengths of a genuine right-angled triangle — no measuring tape ever gives a fraction. The smallest and most famous example is $(3,4,5)$ : indeed $3^2+4^2=9+16=25=5^2$ . This is the same rope-and-knots triangle that ancient builders in both India and Egypt used to lay out perfect right angles long before algebra existed. Other standard triples worth memorising are

(5,12,13),\quad (8,15,17),\quad (7,24,25),\quad (20,21,29)

Primitive triples. A triple is called primitive if $a$ , $b$ and $c$ share no common factor greater than $1$ — their HCF is exactly $1$ . All five triples listed above are primitive. A triple that is not primitive, such as $(6,8,10)$ , is simply a scaled copy of a smaller primitive triple.

Scaling rule. If $(a,b,c)$ is a Pythagorean triple, then for any positive whole number $k$ , the triple $(ka,kb,kc)$ is also a valid Pythagorean triple, because

(ka)^2+(kb)^2=k^2a^2+k^2b^2=k^2(a^2+b^2)=k^2c^2=(kc)^2

where: $k$ is the scale factor. From $(3,4,5)$ , scaling by $k=2,3,4$ gives $(6,8,10)$ , $(9,12,15)$ , $(12,16,20)$ — an infinite family, all describing similar (same-shaped, differently sized) right triangles. Every Pythagorean triple is either primitive or a whole-number multiple of some primitive triple, so the primitive list is genuinely the complete "seed" set. Recognising a handful of standard triples by sight is a genuinely practical shortcut. The moment a triangle's sides look like $6,8,10$ or $9,12,15$ , you can tell instantly — without any lengthy calculation — that it must be right-angled, because these are simply scaled versions of the familiar $3,4,5$ triple. This quick-recognition skill is one of the most useful habits to carry forward from this topic, especially in multiple-choice or time-limited settings, where re-deriving a triple from scratch under time pressure wastes precious minutes. The figure below shows how one primitive triple sits at the root of a whole tree of scaled multiples, while the generating formula (met in the derivation next) reaches sideways to produce entirely new primitive roots.

Figure

A branching diagram showing the primitive triple 3 4 5 at the root with arrows to its scaled multiples 6 8 10 and 9 12 15 and 12 16 20, next to a separate root labelled 5 12 13 produced by the generating formula with m equals 2 and n equals 1

Derivation AThe generating formula always produces a Pythagorean triple

Show, using only algebraic expansion that an eighth-grader already owns, that for any two positive whole numbers $m>n>0$ , the three quantities $a=m^2-n^2$ , $b=2mn$ , $c=m^2+n^2$ always satisfy $a^2+b^2=c^2$ exactly — so the formula is guaranteed to hand out a genuine Pythagorean triple no matter which $m$ and $n$ are picked, without ever needing to test the result by trial.

Step 1. Write down the three expressions to be tested: $a=m^2-n^2$ , $b=2mn$ , $c=m^2+n^2$ . The goal is to show $a^2+b^2=c^2$ by expanding both sides separately and comparing.

Step 2. Expand $a^2$ . Squaring a difference, $a^2=(m^2-n^2)^2=m^4-2m^2n^2+n^4$ .

Step 3. Expand $b^2$ . $b^2=(2mn)^2=4m^2n^2$ .

Step 4. Add $a^2+b^2$ . $a^2+b^2 = (m^4-2m^2n^2+n^4)+4m^2n^2 = m^4+2m^2n^2+n^4$ . The two middle terms combine because $-2m^2n^2+4m^2n^2=2m^2n^2$ .

Step 5. Expand $c^2$ and compare. Squaring a sum, $c^2=(m^2+n^2)^2=m^4+2m^2n^2+n^4$ . This is identical, term for term, to the expression obtained for $a^2+b^2$ in Step 4. Hence $a^2+b^2=c^2$ for every choice of $m>n>0$ , which proves the formula always yields a valid Pythagorean triple. Check: $m=2,n=1$ gives $(3,4,5)$ ; $m=3,n=2$ gives $(5,12,13)$ ; $m=4,n=1$ gives $(15,8,17)$ , the same triple as $(8,15,17)$ with legs swapped.

Step 6. Notice what the derivation did not need. Nowhere in Steps 2–5 did the argument require $m$ and $n$ to be coprime, or to avoid being both odd, or to satisfy any condition beyond $m>n>0$ . The identity $a^2+b^2=c^2$ is a piece of pure algebra that holds for every such pair. What those extra conditions control is a different question — primitivity of the output, not its validity — which is why the formula is such a reliable triple-factory: plug in any two whole numbers with $m>n$ , and a genuine right-triangle triple falls out every time.

Worked example

Problem: Verify that $(8,15,17)$ is a Pythagorean triple and check whether it is primitive; then use the generating formula with $m=4,n=1$ to confirm it is produced by the same formula, and separately verify the scaled triple $(24,45,51)$ .

Step 1. Identify $c=17$ (the largest number), legs $a=8,b=15$ . Compute $a^2+b^2=64+225=289$ and $c^2=17^2=289$ . Since $289=289$ , $(8,15,17)$ is a valid triple.

Step 2. Check primitivity. Factors of $8$ : $1,2,4,8$ ; of $15$ : $1,3,5,15$ ; of $17$ : $1,17$ (prime). The only common factor is $1$ , so $(8,15,17)$ is primitive.

Step 3. Confirm via the generating formula. With $m=4,n=1$ : $a=m^2-n^2=16-1=15$ , $b=2mn=8$ , $c=m^2+n^2=17$ . This gives $(15,8,17)$ , the same triple as $(8,15,17)$ with the legs listed in the other order — the derivation above guarantees this works.

Step 4. Scale by $k=3$ : $(3\times8,\,3\times15,\,3\times17)=(24,45,51)$ . Verify directly: $24^2+45^2=576+2025=2601$ , and $51^2=2601$ . They match.

Final answer: $(8,15,17)$ is a primitive triple, matches the generating formula at $m=4,n=1$ , and its scaled multiple $(24,45,51)$ (with $k=3$ ) is also a confirmed Pythagorean triple.

Competitive stretch

Using $m=5,n=2$ in the generating formula: $a=25-4=21$ , $b=2(5)(2)=20$ , $c=25+4=29$ , giving $(21,20,29)$ — the same numbers as the standard triple $(20,21,29)$ , just with the legs swapped, since order of $a$ and $b$ never matters for the sum $a^2+b^2$ .

Common traps

Assuming any three whole numbers form a triple: most sets do not satisfy $a^2+b^2=c^2$ ; always compute and compare both sides.
Mislabelling which number is $c$ : the largest number must be tested as the hypotenuse, or the check will fail even for a genuine triple.
Confusing "primitive" with "small": primitivity is about having no common factor, not about size — $(20,21,29)$ is primitive despite being numerically larger than the non-primitive $(6,8,10)$ .
Requiring $m,n$ to satisfy no extra conditions: the formula $a=m^2-n^2,\,b=2mn,\,c=m^2+n^2$ works algebraically for any $m>n>0$ (the derivation makes no other assumption), though restricting to $m,n$ coprime and not both odd is what guarantees the result is primitive rather than a scaled copy.
Forgetting that order of $a$ and $b$ does not matter: $a^2+b^2=c^2$ is symmetric in $a$ and $b$ , so $(8,15,17)$ and $(15,8,17)$ describe the exact same triangle; only $c$ is fixed as the largest number.

Ask the tutor: Want to pick your own $m$ and $n$ and have the tutor check whether your generated triple is primitive?

Why does scaling a triple always work, no matter what k is?

Because squaring distributes over multiplication: $(ka)^2+(kb)^2=k^2a^2+k^2b^2=k^2(a^2+b^2)$ . Since the original triple already satisfies $a^2+b^2=c^2$ , this becomes $k^2c^2=(kc)^2$ , which is exactly the condition needed for $(ka,kb,kc)$ to be a triple too. The proportional stretch keeps every angle the same, so you always land on a similar right triangle, just bigger or smaller.

Do I need to memorise the m, n generating formula for exams?

No — at this stage it is enrichment, not core testable content. What you are expected to do confidently is verify a given triple by checking $a^2+b^2=c^2$ , recall standard triples like $(3,4,5)$ , $(5,12,13)$ and $(8,15,17)$ , and scale a primitive triple by a whole number $k$ . The generating formula is worth understanding because it explains where new triples come from and why they never run out, but you will not be asked to apply it cold in a core-board exam — treat the derivation as insight, not a formula to drill.

If m and n can be any whole numbers, why do people restrict them to be coprime and not both odd?

The algebra in the derivation proves $a^2+b^2=c^2$ for any $m>n>0$ , so the triple is always valid. But if $m$ and $n$ share a common factor, or are both odd, the resulting $(a,b,c)$ turns out to have a common factor too — it is a scaled copy of a smaller primitive triple, not a fresh primitive one. Restricting to coprime $m,n$ not both odd is simply the extra condition that guarantees you land on a new primitive triple rather than rediscovering a multiple of one you already know.

Baudhayana (Pythagorean) triples

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads