My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
October 27th, 2010, 09:07 AM   #1
Newbie
 
Joined: Aug 2010

Posts: 17
Thanks: 0

Diophantine Eq, problem. Help please

Show that the Equation below has no solution in non-zero rational numbers x,y,z;

x^2+y^2= 7z^2.
I am interested in the method of proving, the logical steps to the solution.
WizKid is offline  
 
October 27th, 2010, 11:45 AM   #2
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 937

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: Diophantine Eq, problem. Help please

Work mod 4.
CRGreathouse is offline  
October 28th, 2010, 08:48 AM   #3
Newbie
 
Joined: Aug 2010

Posts: 17
Thanks: 0

Re: Diophantine Eq, problem. Help please

Prove by contradiction worked fine for me...

Let n be the least common denominator of x, y, z, so that a, b, c and n are integers, then
(a/n)^2+(b/n)^2 = 7(c/n)^2...by multiplying n^2, we have a^2+b^2 = 7c^2.
If a, b, and c have common factor m, then we can replace them by a/m, b/m, and c/m.
Suppose a, b, and c have no common factor, then one can reduce the last equation by modulo 7
Call A and B the reductions of Modulo 7. The right side is a multiple of 7 and so it becomes 0, and we are left with
A+B = 0 and is satisfied only by the trivial zero, A = 0 and B = 0.
But saying A = 0 and B = 0 is the same as saying that A and B are both multiples of 7 and A^2 and B^2 are multiples of 49 and their sum
7c^2 is a multiple of 49 and Thus c^2 is multiple of 7 which in turn indicates that c is multiple of 7.
And contrary to our assumption, a, b and c share a common factor..>voilą
There is no solution consisting of non-zero rational numbers.

Reduction of Module 4, let me try that and see if it works. But is the above reasoning correct and elegant?
WizKid is offline  
October 28th, 2010, 03:55 PM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 18,166
Thanks: 1425

What if A + B = 7?
skipjack is offline  
October 28th, 2010, 11:20 PM   #5
Newbie
 
Joined: Aug 2010

Posts: 17
Thanks: 0

Re: Diophantine Eq, problem. Help please

Call A and B the reductions of Modulo 7. Another variables just like a and b.
WizKid is offline  
October 29th, 2010, 02:30 AM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 18,166
Thanks: 1425

A + B = 7 doesn't imply that A = B = 0. You need to show that A and B cannot be 1 and 6 or 2 and 5 or 3 and 4.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
diophantine, problem



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Diophantine Fun mathbalarka Number Theory 3 October 11th, 2012 09:24 AM
Another logarithmic diophantine problem !!! mathbalarka Algebra 12 September 23rd, 2012 05:01 AM
diophantine x³ + 8x² - 6x + 8 = y³ Pell's fish Number Theory 9 March 23rd, 2012 12:07 AM
Diophantine Problem johnsy123 Number Theory 2 March 14th, 2012 06:10 AM
Diophantine Fun mathbalarka Algebra 1 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.