2

I've tried using modulo $3$, and I get it down to

$y^2 + z^2 = 0 \pmod 3$ ; I don't know where to go from here though.

I justified my answer by stating that, because we're in $\pmod 3$ and we need non-trivial solutions, the only solutions possible are $y = 1,2$ and $z = 1,2$; all of which will not give us $0$.

Thanks!

DeepSea
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Lerbi
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  • Yeah, exactly. I was talking to my teacher (who is a very great teacher, by the way), but he was being rather vague. The hint he gave me was "what do x, y, and z, have in common, common factor-wise?" – Lerbi Nov 14 '14 at 01:45
  • Try modulo $4$. – Thomas Andrews Nov 14 '14 at 01:48
  • What's this? You kidding? In one subject prove that no solutions. http://math.stackexchange.com/questions/1001787/does-the-equation-has-a-non-trivial-solution In this thread looking for a solution. In General, the formula for this equation there. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Nov 14 '14 at 04:49

2 Answers2

4

Hint: $y^2+z^2 = 0\pmod 3 \to y=z=0\pmod 3$

DeepSea
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  • Sorry, I'm fairly new to modular arithmetic. Can you explain your statement? – Lerbi Nov 14 '14 at 01:44
  • It means if $3\mid y^2+z^2$ then $3\mid y$, and $3\mid z$ – DeepSea Nov 14 '14 at 01:45
  • So, now I'm confused again. That means we can choose any values of y and z to satisfy this? For instance, we could choose y = 6 and z = 9? Or are we restricted to 0, 1, and 2? In the former case, nothing is really being proved, is it? – Lerbi Nov 14 '14 at 01:45
  • ...and you should convince yourself that that statement is true – Simon S Nov 14 '14 at 01:47
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Modulo $4$, this equation is $3x^2+3y^2+3z^2\equiv 0\pmod 4$, or $x^2+y^2+z^2\equiv 0\pmod 4$. The only squares modulo $4$ are $0$ and $1$, so conclude that $x,y,z$ must all be even.

Thomas Andrews
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  • Okay, so knowing that x, y, and z must all be even, how does this prove that no integer solution satisfies the equation? – Lerbi Nov 14 '14 at 01:55
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    If $x,y,z$ have a common factor, $d$, then $x/d,y/d,z/d$ is also a solution. So there must be a solution where $x,y,z$ do not have a common factor. – Thomas Andrews Nov 14 '14 at 02:06
  • Awesome, thanks Thomas. I was tardy to thank you, but you saved my butt! – Lerbi Nov 21 '14 at 22:40