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Pythagorean triples

Every primitive Pythagorean triple (integer sides with $\gcd(a,b,c)=1$ ) can be generated from a pair of integers $m > n > 0$ , coprime and of opposite parity:

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

Every other Pythagorean triple is just an integer multiple of a primitive one — this is why triples like 6-8-10 and 9-12-15 are all built from the primitive 3-4-5.

Sums of two squares

A key factoring identity: $a^2-b^2=(a+b)(a-b)$ turns a difference of squares into a product, which is often the entry point for these problems. A deeper fact governs which integers can be written as a sum of two squares: a positive integer $n$ is expressible as a sum of two squares if and only if every prime factor of $n$ that is $\equiv 3 \pmod4$ appears to an even power.

Worked example

Problem: Find all primitive Pythagorean triples with hypotenuse 25.

Solution: We need $m^2+n^2=25$ with $m>n>0$ , coprime, opposite parity. Testing: $m=4, n=3$ gives $16+9=25$ ✓ (coprime, opposite parity).

This generates the triple:

a=m^2-n^2=16-9=7, \qquad b=2mn=24, \qquad c=25

So the triple is $(7, 24, 25)$ — and this is the only primitive triple with hypotenuse 25 (the other decomposition, $25=5^2+0^2$ , fails since it requires $n>0$ ).

Practice problems

1.Using $m=3, n=2$ , generate a primitive Pythagorean triple.

2.Is 21 expressible as a sum of two squares? ( $21=3\times7$ )

3.Is 45 expressible as a sum of two squares? ( $45=3^2\times5$ )

4.Solve $x^2-y^2=15$ in positive integers, using the factoring identity.

Quadratic Diophantine Equations

Pythagorean triples

Sums of two squares

Practice problems