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.

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.)

Yes, it is possible for each of the sides of a right triangle to have odd integer lengths. To demonstrate this, let's consider a specific example:

Suppose we have a right triangle T with side lengths a = 3, b = 4, and c = 5. Here, all three sides are odd integers.

To verify that this triangle is indeed a right triangle (satisfying the Pythagorean theorem), we can apply the theorem by squaring the lengths of the two smaller sides and adding them together. If the sum of the squares of the two smaller sides is equal to the square of the longest side, then it is a right triangle.

For our example, we have a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25, and c^2 = 5^2 = 25. Since a^2 + b^2 = c^2, this verifies that the triangle T is a right triangle.

Therefore, we have demonstrated that it is indeed possible for each side of a right triangle to have odd integer lengths.