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Show that there are no intergers $x$ and $y$ such that


Hint from professor:

Consider the equation in a convenient $\mod (n)$ so that you end up with a polynomial in a single variable. Then proceed as solving number of congruence.

Im not sure how to approach this question

Since $P(x,y)=x^2-5y^2=2$

then $x^2-5y^2=0$ $\to$ $x^2=5y^2$

we have $5y^2\equiv0\mod(x)$

then how do I continue..?

Thank you!!

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up vote 3 down vote accepted

Suppose there exists integers $x$ and $y$ such that $x^2 - 5y^2 = 2$. Use the fact that

Every square number is congruent to either $0$ or $1$ modulo $4$. $(\ast)$

Hence, $x^2 - 5y^2 \equiv x^2 - y^2 \equiv 2 \pmod{4}$. However, the difference of two squares $x^2 - y^2 \equiv -1, 0, 1 \pmod{4}$ due to $(\ast)$.

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But that doesn't follow the professor's hint (which gives a simpler proof). – Math Gems Feb 12 '13 at 6:44
I agree, now that I saw @awllower's answer, I think that follows the hint better. And I like that answer better :) – Herng Yi Feb 12 '13 at 6:45
It's even simpler since reciprocity is overkill here - see my answer. – Math Gems Feb 12 '13 at 6:46
I think simplifying to one variable is a powerful stroke, and testing the 5 cases is really easy anyway. Matter of taste? haha... – Herng Yi Feb 12 '13 at 6:50

I think your professor means to divide the equation by $5$, so that it becomes $x²\equiv 2 \pmod 5$. But, by the supplementary law of quadratic reciprocity, this is impossible.

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Judged by what your professor said, especially the single-variable part, I guess you are already familiar with quadratic reciprocity laws. Even if you are not, you can check five cases $\pmod5$ to see that there is no solution to the reduced equation. – awllower Feb 12 '13 at 6:24

Note that, every perfect square is either $0$ or $1$ modulo $4$. Now check all the cases to see that $x^2-5y^2$ is never equal to $2$ modulo $4$.

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Following the hint: $\rm\ mod\ 5\!:\ x^2\! - 5y^2 \equiv x^2\in \{0, \pm1, \pm 2\}^2\! \equiv \{0, 1, 4\},\:$ so is not $\equiv 2.$

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