Solving Diophantine Equations Involving GCD and Divisibility

  • Thread starter Thread starter bronxbombas
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving that if gcd(a, b) = 1 and both a and b divide n, then ab also divides n. The key equation used is ax + by = 1, where x and y are integers. The solution involves multiplying this equation by n, leading to the expression n = nax + nby, which confirms that both a and b divide n. Additionally, a secondary problem regarding the oddness of (2n)!/(2^n*n!) for nonnegative integers n is mentioned, suggesting an expansion approach for verification.

PREREQUISITES
  • Understanding of GCD (Greatest Common Divisor) and its properties
  • Familiarity with Diophantine equations
  • Basic knowledge of factorials and their properties
  • Experience with integer algebra and divisibility rules
NEXT STEPS
  • Study the properties of GCD and its implications in number theory
  • Learn about Diophantine equations and methods for solving them
  • Explore the concept of factorials and their applications in combinatorics
  • Investigate the proof techniques for divisibility in integer sets
USEFUL FOR

Mathematicians, students studying number theory, and anyone interested in solving Diophantine equations and understanding divisibility concepts.

bronxbombas
Messages
9
Reaction score
0

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.
 
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K