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 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 = 2-z 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. Thanks from topsquark 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: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra Individually the equations obviously have integer solutions, just pick $x$ and $y$ and plug in to determine $z$. Thanks from topsquark
 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 $(x^2+y^2-9z)=3$, 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. Thanks from Joppy
November 12th, 2017, 08:31 PM   #6
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 Originally Posted by johng40 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 $(x^2+y^2-9z)=3$, 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.
Thank you I solved it using your first problem solution.

November 12th, 2017, 08:57 PM   #7
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 Originally Posted by v8archie . . . just pick $x$ and $y$ and plug in to determine $z$.
Doing that gives a value for $z$ that isn't an integer.

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