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$n=3 \pmod 4$ a Natural number, prove there are no Integer solutions to $X^2-nY^2=-1$

I don't really know how to start this one...a little help?

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What numbers are squares mod 4? –  jspecter Feb 6 '13 at 17:38

3 Answers 3

Hint Look at $X^2-nY^2=-1$ modulo 4.

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If there is a sulotion to an equation, then there is a sulotion to this equation mod some m? –  user1932595 Apr 4 '13 at 6:47

HINT: the only squares modulo $4$ are $\bar 0$ and $\bar 1$. On the other hand the equation $X^2-nY^2=-1$ reduces modulo $4$ to $$ X^2+Y^2=\bar 3 $$ since $\bar n=\overline{-1}$.

Thus ....

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And thus, it dawns on me that you might be right :P –  Rachel Bernoulli Feb 6 '13 at 17:53

An alternative way:
If it has a solution, then $-1$ is a quadratic residue modulo $n$. But $n \equiv 3 \pmod4$, hence this is not possible by the supplementary laws of quadratic reciprocity.

EDIT
Since $n\equiv 3 \pmod4$, there is some prime divisor of $n$ which is $\equiv 3 \pmod4$, hence justifying our use of quadratic reciprocity.

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One might consider the use of quadratic reciprocity on such an elementary question to be quite unnecessary, but it makes me feel more systematic however. Of course elementary considerations by division by $4$ suffuces to prove it, we can, however, regard this equation as stating that $-1$ is a residue of $n$, thus giving an (unnecesssary) application of the law. Without a doubt, the use of reducing modulo $4$ is likewise systematic in some sense. But that is another story.^^ –  awllower Feb 7 '13 at 10:33

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