Infinite recipe
Euclid’s formula with \(m>n\), opposite parity, and \(\gcd(m,n)=1\) spits out every primitive triple—proof there are infinitely many right triangles with all-integer sides.
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A Pythagorean triple is a set of positive integers \((a,b,c)\) satisfying \(a^2+b^2=c^2\). These numbers measure the sides of a right triangle, with \(c\) as the hypotenuse. The classic example is \((3,4,5)\), since \(3^2+4^2=9+16=25=5^2\). Triples come in two flavors: primitive (no common factor: \(\gcd(a,b,c)=1\)) and non-primitive (multiples of a primitive triple). For instance, \((6,8,10)\) is a non-primitive multiple of \((3,4,5)\).
All primitive triples can be produced by Euclid’s formula. Choose integers \(m>n\) with \(\gcd(m,n)=1\) and exactly one of \(m,n\) even, then \[ a=m^2-n^2,\quad b=2mn,\quad c=m^2+n^2. \] Every primitive triple appears exactly once this way. Multiplying by a positive integer \(k\) yields every non-primitive triple \( (ka, kb, kc)\).
In this generator, set a max hypotenuse to constrain size. Turn on “Primitive only” to explore the fundamental families, or include multiples to see all triples up to your bound. You can also search for triples containing a specific leg (e.g., all triples with \(a=20\)) or a specific hypotenuse (e.g., \(c=65\)). The table can be sorted by \(a\), \(b\), \(c\), perimeter, or aspect ratio \(a{:}b\), and you can export the results to CSV for class handouts or further analysis. Everything runs locally in your browser—no uploads—so it’s fast and private.
Euclid’s formula with \(m>n\), opposite parity, and \(\gcd(m,n)=1\) spits out every primitive triple—proof there are infinitely many right triangles with all-integer sides.
\((3,4,5)\) is the lone triple of three consecutive integers. Any other triple must skip a number—an easy way to spot impostors.
Pick any odd \(k>1\); \((k,\tfrac{k^2-1}{2},\tfrac{k^2+1}{2})\) is a valid triple. Example: \(k=99\) yields \((99,4900,4901)\).
Every primitive triple’s area is divisible by 6. One leg is a multiple of 4, one is a multiple of 3, so \(A=\tfrac{ab}{2}\) always keeps a factor of 3.
Primitive triples correspond to reduced rational points on the unit circle via \((\tfrac{m^2-n^2}{m^2+n^2}, \tfrac{2mn}{m^2+n^2})\). Counting lattice right triangles = counting rationals on a circle.