Solving Diophantine Equations Involving GCD and Divisibility

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Homework Statement


Suppose that gcd(a,b)=1 and that a|n and b|n. Prove that ab|n.


Homework Equations


Since we know that gcd(a,b)=1, we can say that ax+by=1 for some x,y as elements of the integer set.


The Attempt at a Solution


My professor said I should multiply the entire equation by n, but I still can't figure it out. Any help would be appreciated. Thanks in advance.
 
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I also have another problem that takes priority over this one if anybody can help.

Prove that (2n)!/(2^n*n!) is an odd number when n is a nonnegative integer.
 
If you multiply the equation by n, you get n=nax+nby. Now use the fact that both a & b divide n.

For your second problem, have you tried expanding (2n)!/(2^n*n!) out?
 
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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