Either meaning that n is even or meaning that n is odd. So assume that n is equivalent to 1. Then n^3=1*1*1=1. Thus .Originally Posted by polish13
-Dan
Questions
Prove or give counter examples for the following problems
1) For any integer n, (n^2 + 2n)mod4 is equal to 0 or 3.
2) Suppose m, n, and d are integers, and m mod d = n mod d. Prove that dl(m - n) (says d divides (m minus n).
3) If n^3 is odd, then n is odd
4) If a, b, and c ate integers and a^2 + b^2 = c^2, then at least one of a, b, and c is odd
5) Prove that the cube root of 4 is irrational
Please help
Either n is even or it is odd. If n is even then n is equivalent to 0 or 2 (mod 4). In the first case n^2=0*0=0, and 2n=2*0=0. So n^2+2n is equivalent to 0+0=0 (mod 4). In the second case n^2=2*2=4->0 (mod 4) and 2n=2*2=4->0 (mod 4), so n^2+2n is equivalent to 0+0 (mod 4).Originally Posted by polish13
Now assume n is odd. So n is equivalent to 1 or 3 (mod 4). In the first case n^2=1*1=1 and 2n=2*1=2. So n^2+2n=1+2=3 (mod 4). In the second case n^2=3*3=9->1 (mod 4) and 2n=2*3=6->2 (mod 4). So n^2+2n=1+2=3 (mod 4).
-Dan
Using my previous two posts, you should find it easy to say that if n is even then n^2 is even and if n is odd then n^2 is odd.Originally Posted by polish13
So all we need to consider is if a^2, b^2, or c^2 are even or odd. Let's take the (essentially three) cases:
1)a^2, b^2 are both even
2)a^2 is even, b^2 is odd (equivalent to a^2 is odd, b^2 is even)
3)a^2, b^2 are both odd.
Let's take 1) last.
2) a^2+b^2 is even+odd = odd. So c^2 is odd.
3) a^2+b^2 is odd+odd = even. So c^2 is even. Note that we already have two odds in the problem so we are done.
1) a^2+b^2 is even+even = even. So c^2 is EVEN! This implies that c is even. This means that your statement is FALSE! (Example: a=6, b=8, c=10).
-Dan
I'm not sure of which method you need to use here. I can't say I know of one using modular Mathematics because we're dealing with the possibility of rational solutions to the equation . However, there is a simple proof using the Rational Roots test:Originally Posted by polish13
If has a rational root it will be of the form: . As none of these possibilities work, x has no rational solution. Thus is irrational.
-Dan
Based on Euclid's proof,Originally Posted by polish13
Assume that,
Then,
Thus,
Notice when you use the fundamental theorem of arithmetic the right hand has a number of prime factors not divisible by 3, while the left hand side does. Thus, an impossibility because of the uniquess of factorization into primes.