Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$n=3 \pmod 4$ 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?

share|cite|improve this question
What numbers are squares mod 4? – jspecter Feb 6 '13 at 17:38

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

share|cite|improve this answer
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 ....

share|cite|improve this answer
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.

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

share|cite|improve this answer
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

Your Answer


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.