24 May 12, 04:54PM
(This post was last modified: 24 May 12, 05:16PM by Roflcopter.)
(24 May 12, 04:39PM)Habluka Wrote: 1. Pythagorean Theorem states that a^2 + b^2 = c^2 and a Pythagorean triplet is 3 numbers that satisfy this that are positive integers. a and b must therefore be positive integers and multiplying and adding integers will always get another integer, so you know that a * a + b * b will be an integer, which means c is the square root of an integer. Because of the definition of a Pythagorean triplet, this means c is a perfect square and thus is unable to be prime.
c is not necessarily a perfect square though. For example the triple (3,4,5) is Pythagorean (since 9+16=25) and c is in fact prime.
(24 May 12, 04:39PM)Habluka Wrote: 2. there are no conditions given that state that x < x', meaning that x can be greater than x'. In this condition, there would be no way that x < a/b < x'. Rewrite this one and I don't like proofs by construction.
You're right it was a bit unclear. You should take x < x'.
(24 May 12, 04:39PM)Habluka Wrote: 3. 10^6 - 1 (takes no programming or real calculations since sqrt(10^12 = 10^6)
A squarefree number is a positive integer n such that no square divides n wholly. For example 18 is not squarefree since 3 squared divides it. But 30 is since it has no square factors.
But that answer is far off. Try working with smaller versions of the problem first. For instance there are 11 squarefrees less than 16: 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15.