if the sides of a rightangle triangle are integral in their lowest form show only one side is divisible by 5?

Answer 1

Do you want us to show that one and only one side is divisible by 5?


Or just show that only one is divisible by 5 (your question)?
To answer your question:

Assume that 2 sides are divisible by 5. Let these sides be a and b. Let's also assume that a≥b (this can be done without losing generality). We don't know if a or b is the hypotenuse, but since a≥b we know that for the third side c, c^2=a^2±b^2 (+ if c is the hypotenuse, - is a is the hypotenuse). But a and b are both divisible by 5, so a^2 and b^2 are divisible by 25. Thus c^2 is divisible by 25 and c is divisible by 5. Therefore by dividing a, b, and c by 5, you can find smaller integral lengths.

Answer 2

well let's see... suppose that two of the sides could be divided by 5. Say
a = 5x b = 5y [I will assume a>=b]
According to the Pythagorean theorem, the third side is
c = sqrt (a^2 +/- b^2)
= sqrt (25x^2 +/- 25y^2)
= 5 sqrt (x^2 +/- y^2)
If c is integral, sqrt(x^2 +/- y^2) must be a rational number, therefore an integer, and c would also be a multiple of 5. But then all three sides are multiples of 5 and not in their lowest form.