According to many history sources, several formulas were developed to define certain types of Pythagorean triangles. One set that was attributed to Pythagoras took the form of x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. It was ultimately discovered that the formulas created only triangles where the hypotenuse exceeded the larger leg by one.
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.)
In every primitive Pythagorean triple, x, y and z, either x or y is divisible by 2 or 4, either x or y is divisible by 3, and either x, y, or z is divisible by 5. The area is always divisible by 6 and the product of all three sides is always divisible by 60.
Why?
1--Since either x or y is equal to 2mn or 4mn, the one that is equal to 2mn, or 4nm, is obviously divisible by 2 or 4.
2--If either m or n is divisible by 3, then x or y = 2mn is divisible by 3. If neither m or n is divisible by 3, then y = m^2 - n^2 is as the squares of both m and n are of the form 3k + 1.
3--If the product of the 3 sides is divisible by 5, then one of the sides is divisible by 5. The product is 2mn(m^2 - n^2)(m^2 + n^2) = 2mn(m^4 - n^4). If m or n is a multiple of 5, so is 2mn. If neither m or n is a multipe of 5, then (m^4 - n^4) is as m is then of the form 5k+/- 1 or 5k+/-2. Expanding the fourth power of each of them by the binomial theorem, it is found that m^4 is of the form 5h + 1 and similarly for n^4.
5--With x divisible by 4, x or y divisible by 3 and x, y or z divisible by 5, the area , xy/2, is divisible by 4x4/2 and xyz is divisible by 4x3x5.
(Ref: Mathematical Recreations by Maurice Kraitchik, Dover Publications, Inc., 1953.)
Suppose T is a right triangle whose sides have lengths a, b, and c. Is it possible that each of a, b, and c is an odd integer?
Yes
No
Explain and demonstrate your answer.
1 answer
1 answer