Classify primitive right triangles

Try

The triples (3, 4, 5), (5, 12, 13) and (8, 15, 17) describe integer-sided right triangles. Scaling any one produces more examples, so first separate primitive triangles, whose three sides have greatest common divisor one. The question is not just how to manufacture examples but whether one formula produces every primitive triangle. The equation is x²+y²=z² with positive integers, and exchanging the two legs does not change the triangle.

Predict

Predict the parity pattern before attempting factorization. Two odd legs would make z² congruent to two modulo four, impossible. Two even legs would contradict primitivity. Thus exactly one leg is even and the hypotenuse is odd. Name the odd leg x and the even leg y. The identity (z+x)(z−x)=y² suggests factoring, but both factors are even; dividing each by two removes the common factor that would otherwise obstruct the square argument.

Experiment

Compare parameter pairs and diagnose which generate primitive triples.

m	n	m²−n²	2mn	m²+n²	Primitive?
2	1	3	4	5	Yes
3	2	5	12	13	Yes
3	1	8	6	10	No
4	2	12	16	20	No

Change one condition at a time: same parity in the third row, shared divisor in the fourth. The identity always holds, but primitivity needs both restrictions.

Understand

In a primitive triple, any common divisor of x and z also divides y² and therefore y, so the sides are pairwise coprime. Put A=(z+x)/2 and B=(z−x)/2. A common divisor divides A+B=z and A−B=x, hence A and B are coprime. Their product is (y/2)². Unique factorization now forces each to be a square: write A=m² and B=n² with m>n>0. Subtraction and addition give x=m²−n² and z=m²+n²; positivity gives y=2mn. The parameters are coprime because A, B are. They have opposite parity because m²+n²=z is odd. This proves necessity and explains every hypothesis rather than attaching restrictions after the formula.

Apply

Conversely, the displayed formulas satisfy the right-triangle identity by expansion. For coprime opposite-parity m, n the odd leg is odd, so two cannot divide all sides. If an odd prime divides m²−n² and 2mn, it divides one of m, n and then the other, contradicting coprimality. Thus the triple is primitive. Every nonprimitive triple is a positive integer multiple of a primitive one after dividing by the common gcd. To recover parameters from a given primitive triangle, choose its odd leg and take the square roots of (z+x)/2 and (z−x)/2; this also proves uniqueness with the chosen leg convention.

Continue

Use the classification to design a triangle satisfying an additional condition, such as a prescribed even leg or a perimeter bound. List factor pairs of half the even leg, then filter coprimality and parity instead of blindly searching all three sides. Zero and negative parameters were excluded because this is a classification of positive side lengths. The same formulas will support infinite descent and rational points on a circle, where slope rather than integer factorization provides another route to the parametrization.

Return to the number theory unit to choose a targeted retry. Continue with decide which integers are sums of two squares.