November 11th, 2017, 09:03 PM  #1 
Newbie Joined: Oct 2017 From: sweden Posts: 20 Thanks: 0  Diophantine equations
Can anyone help me solve that these two equations have no integral solutions? $x^2+y^2=9z+3$ $x^2+2y^2=8z+5$ I am very new to Diophantine equations, and it's like the first two problems in this particular chapter in the textbook and I am already stuck. Many helps will be appreciated thank you! Last edited by skipjack; November 12th, 2017 at 08:27 AM. 
November 12th, 2017, 02:15 AM  #2 
Newbie Joined: Nov 2017 From: US Posts: 5 Thanks: 4 
If you subtract equation (1) from (2), then you get y^2 = 2z Assuming you are looking for positive integers, you would see one option is z=1 and y=1 (other option is z=2, y=0). Substituting y=1 and z=1 in either equation, you get x^2=11 ==> no integer solution. Substituting y=0 and z=2 in either equation, you get x^2=21 ==> no integer solution. See how you can extend this logic for negative integers. Last edited by skipjack; November 12th, 2017 at 08:26 AM. 
November 12th, 2017, 08:43 AM  #3 
Newbie Joined: Oct 2017 From: sweden Posts: 20 Thanks: 0 
Sorry for the unclear statement of the problem, actually those two equations are two separate problems, and are not related (Edited).

November 12th, 2017, 08:47 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
Individually the equations obviously have integer solutions, just pick $x$ and $y$ and plug in to determine $z$.

November 12th, 2017, 02:37 PM  #5 
Member Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 
I don't understand v8archie's reply at all. Here's the first one: Assume x, y and z are solutions. Suppose 3 divides x. Then 3 also divides y and so 9 divides $(x^2+y^29z)=3$, a contradiction. Symmetrically, 3 is prime to y. Then mod 3, $x^2+y^2=2\neq 0$, the final contradiction. I hope you can do the second one; if not, I'll try to think of a solution. 
November 12th, 2017, 08:31 PM  #6  
Newbie Joined: Oct 2017 From: sweden Posts: 20 Thanks: 0  Quote:
 
November 12th, 2017, 08:57 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,065 Thanks: 1621  

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