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 October 10th, 2012, 05:15 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Diophantine Fun This one's pretty challenging but not as hard as we think : Prove that $a^2 + b^2= 3c^2$ has no solution in positive integers Hint : Consider the equation modulo 4 Balarka .
 October 10th, 2012, 06:46 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 Fun Pretty easy once you divide out common factors.
 October 10th, 2012, 07:06 AM #3 Global Moderator   Joined: Dec 2006 Posts: 18,595 Thanks: 1493 If 3 divides a² + b², both a and b must be divisible by 3, so 9 divides a² + b², which implies c is divisible by 3. Hence one could divide all three variables by 3 to get another solution in integers. After sufficient repetition, one must get a = b = c = 0, so this is true to start with, showing that there is no solution in positive integers.
October 11th, 2012, 09:24 AM   #4
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Re: Diophantine Fun

Quote:
 Originally Posted by mathbalarka Prove that $a^2 + b^2= 3c^2$ has no solution in positive integers
Geesh, if Big Pete Pythagoras looked at this quickly, he'd get a heart attack

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