Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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!!

share|improve this question
add comment

4 Answers

up vote 2 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 $(\ast)$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
add comment

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 $4$.

share|improve this answer
add comment

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 $\equiv 2.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.