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Prove $x^2 + 2y^2 \neq 805, x,y\in\mathbb{Z}$.

We did this in class and, for the life of me, I cannot remember how to finish the problem.

It starts out by taking all of the values to be $\mod5$.

So, $[x^2]_5 + 2[y^2]_5 \neq [805]_5 \neq [0]_5$.

Naturally (heh), the possible values of $[x^2]_5$ are $[0]_5$, $[1]_5$, or $[4]_5$ and it follows for $2[y^2]_5$ the values are $[0]_5$, $[2]_5$, or $[3]_5$.

Thus, all combinations of $[x^2]_5\neq[0]_5\neq2[y^2]_5$, the equation equals everthing BUT $[0]_5$. Which is good.

However, then we reach a point where we must ask ourselves why it is impossible in the equation for $[x]_5$ and $[y]_5$ to be congruent to $[0]_5$. This is the part I don't remember how to do: How do I prove that $[0]_5 + [0]_5 \neq [0]_5$ is a contradiction/does not apply for $x^2 + 2y^2 = 805$?

I do remember it being a rather simple contradiction, but I cannot think of it.

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    It's a proof based math course that I have a test in tomorrow. I was reviewing and came across this problem and got stuck. I knew I could have done an exhaustive search at this point, but that's not very good for taking a test.2012-11-29

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Hint: If $x^2$ and $y^2$ are divisible by $5$, then so are $x$ and $y$, and $x^2$ and $y^2$ are then divisible by $25$.

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    Of course. Thank you. I would vote up if I could. :)2012-11-29