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November 11th, 2017, 10:03 PM   #1
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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 09:27 AM.
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November 12th, 2017, 03:15 AM   #2
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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.
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Last edited by skipjack; November 12th, 2017 at 09:26 AM.
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November 12th, 2017, 09:43 AM   #3
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Sorry for the unclear statement of the problem, actually those two equations are two separate problems, and are not related (Edited).
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November 12th, 2017, 09:47 AM   #4
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Individually the equations obviously have integer solutions, just pick $x$ and $y$ and plug in to determine $z$.
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November 12th, 2017, 03:37 PM   #5
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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.
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November 12th, 2017, 09:31 PM   #6
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Quote:
Originally Posted by johng40 View Post
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.
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November 12th, 2017, 09:57 PM   #7
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Quote:
Originally Posted by v8archie View Post
. . . 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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