The side lengths of a right triangle are each an integral number of units. If one of the legs is 13 units, what is the perimeter of the triangle?


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  1. Look for three integers that satisfy
    a^2 + b^2 = c^2

    Have you tried 5, 12, 13? That is the only possibility with a 13-unit side.

    Add them up for the perimeter.

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  2. All Pythagorean Triples of the form x^2 + y^2 = z^2 derive from x = k(m^2 - n^2), y = k(2mn), and z = k(m^2 + n^2) where k is any positive integer and m and n are arbitrary positive integers, m greater than n. Pythagorean Triples that have no common factor, or a greatest common divisor of 1, are called primitive. Those with a common factor other than 1 are called non-primitive triples. Primitive Pythagorean Triples are obtained only when k = 1, m and n are of opposite parity (one odd one even) and have no common factor, and m is greater than n. (For x, y, & z to be a primitive solution, m and n cannot have common factors and cannot both be even or odd. Violation of these two limitations will produce non-primitive Pythagorean Triples.)

    Assuming k = 1, and that your statement "If one of the legs is 13 units" means one of the legs at 90º to one another is 13 units, then 13 is the side which derives from m^2 - n^2. Quick observation shows that m must be 7 and n must be 6 making 13 = 7^2 - 6^2. With m = 7 and n = 6, the three sides become a = 7^2 - 6^2 = 13, b = 2(7)6 = 84 and c = 7^2 + 6^2 = 85.

    Since the perfect squares are the sequential sum of the odd integers starting with 1, the sum of the 2nd and 3rd squares, 4 + 9 = 13 deriving from m = 3 and n = 2 which leads to another Pythagorean triangle where a = 3^2 - 2^2 = 5, b = 2(3)2 = 12 and c = 3^2 + 2^2 = 13.

    No values of m an n can yield a side of 2mn as it always results in an even number.

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