can the sides of a right triangle be three consecutive odd integers?

Answer 1

No. The sets of integers that can be sides of a right triangle are called Pythagorean triples. These are triples of integers (a,b,c) for which a^2 + b^2 = c^2. Pythagorean triples have the property that exactly two of them are odd, and one even. That right there tells you that they cannot be three consecutive odd integers. There is a formula for generating all Pythagorean triples (that are coprime). If m and n are positive integers with m > n, then

a = m^2 - n^2, b = 2mn, c = m^2 + n^2

is a Pythagorean triple. From this formula you should be able to prove directly that a, b, and c can never be three consecutive odd integers.

Answer 2

Using the Pythagorean Theorem to answer this question...

a² + b² = h²

a² + (a+2)² = (a+4)²

a² + a² + 4a + 4 = a² + 8a + 16

a² - 4a -12 = 0

( a - 6 ) ( a + 2 ) = 0

The only positive factor is 6, so the ONLY right triangle formed by three consecutive odd or even integers is the 6-8-10 triangle.

Answer 3

No.

1. The square of an odd number is always odd.
2. The sum of two odd numbers is always even.
3. The square root of an even number is never odd, because from 1: the square of an odd number is always odd.

Thus, the hypotenuse will never be odd if the two sides are odd.

Answer 4

No. What you're describing is x^2 + (x+2)^2 = (x+4)^2
The only solutions to that are x = 6 or -2.

Answer 5

Pythagorean triples - check the websites. If you find any tell me!

Answer 6

Yup, what Iheart80 said

Answer 7

Here is very, very careful explanation
Hope it helps
:-)