Proving: If a is Even, Then 4 Divides a

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Homework Statement


Let a and b be a natural numbers such that a2 = b3. Prove the following proposition:
If a is even, then 4 divides a.


Homework Equations


Definition: A nonzero integer m divides an integer n provided that there is an integer q such that n = m * q.

Definition: A even number m can be represented by the relationship m = 2 * n where n is an integer.


The Attempt at a Solution


Let a = 2n where b is any integer. Let b = 2m where m is any integer (from another theorem, the cube of any even is even).

a^2 = b^3
(2n)(2n) = 8m^3
2n = 4m^3/n

I am not even sure if I did all of the above correctly; but, this is as far as I got.
 
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Try to show: if 2 divides b3, then 8 divides b3.
 
What you have done is pretty good up to here...

2n = 4m^3/n

You have already shown that

4n^{2} = 8m^3

n^{2} = 2m^3

This imples n is even, do you see why ?

If n is even then 2n is definitely divisible by 4.
 
Last edited:
╔(σ_σ)╝ said:
What you have done is pretty good up to here...



You have already shown that

4n^{2} = 8m^3

n^{2} = 2m^3

This imples n is even, do you see why ?

If n is even then 2n is definitely divisible by 4.

Wow... thank you! I finally see how it is done.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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