
Abstract Algebra Abstract Algebra Math Forum 
 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 nonzero 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. 
October 27th, 2010, 11:45 AM  #2 
Global Moderator 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.

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 nonzero rational numbers. Reduction of Module 4, let me try that and see if it works. But is the above reasoning correct and elegant? 
October 28th, 2010, 03:55 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,166 Thanks: 1425 
What if A + B = 7?

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.

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.


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 