Show that the following equations have no solutions in $\mathbb{Z}$

Question

I want to show that $$x^2 - n\, y^2 = m$$ does not have any solutions for $x,y\in \mathbf{Z}$ for the pairs $(n,m) \in \{(2,3), (3,5), (6,11) \}$ by considering the above equation modulo carefully chosen primes. But I do not understand what is the logic for why this could be useful. I could try, for example when $(n,m) = (2,3)$, that modding out by 2 reveals $x^2$ must be odd and hence that $x$ must be odd. But I don't see where to go from there. Moreover, I don't understand what is the structure of the problem that suggests that modding out by small primes should furnish the result.

Answer 1

Extra note: similar problems will not always be solvable by just checking a few primes. For example, there is no immediate way to show that $x^2 - 37 y^2 = 3$ is impossible with integers $x,y.$

ORIGINAL: Discriminant of $x^2 - n y^2$ is $4n.$

Legendre symbols $$ (8|3) = (2|3) = -1. $$ $$ (12|5) = (3|5) = (5|3) = (2|3) = -1. $$ $$ (24|11) = (6|11) = (2|11)(3|11) = -(3|11) $$ because $11 \equiv 3 \pmod 8,$ so $2$ is not a quadratic residue. $$ (24|11) = -(3|11) = (11|3) = (2|3) = -1 $$

Prime divisors of $k^2+(k+1)^2$

Answer 2

For $(n,m)=(2,3)$ you can look at the equation mod $8$. Squares are only congruent to $0, 1$ and $4$ modulo $8$ and, as you said, $x$ must be odd, so $x^2$ is congruent to $1$ mod $8$. If $y^2$ was congruent to $0, 1$ or $4$, then $x^2-2y^2$ would be congruent to $1, 7$ or $1$ mod $8$ but not $3$. In conclusion, there can't be a solution for the equation in the integers.

Now consider $(n, m)=(3,5) $ or $(6,11) $. Since $3$ and $6$ are both divisible by $3$, $x^2$ must equal $m $ modulo $3$. But the only quadratic residues mod $3$ are $0$ and $1$ while $5$ and $11$ are congruent to $2$ modulo $3$. Hence, there can't be any solutions to the equations in the integers.

Answer 3

First $m^2x^2-nm^2y^2=m\Rightarrow m(x^2-ny^2)=1$ impossible because $\dfrac1m\notin\mathbb Z$.

Second $\mathbb F_3^2=\{1,0\},\space \mathbb F_{5}^2=\{1,4,0\},\space \mathbb F_{11}^2=\{1,4,9,5,3,0\}$.

In the three cases $2$ is not square,but in the same three cases one has $2$ as the inverse of $2,3,6$ so we have

$$x^2-ny^2=m\Rightarrow2x^2=y^2\text{ in }\mathbb F_m\text{ for } m=3,5,11$$ This prove the required impossibility.

NOTE.-When $m$ is a prime and not small, if $n$ has $2$ as inverse modulo $m$ then the equation $x^2-ny^2=m$ has no solution if $2$ is not a square modulo $m$ and this can be obtained by reciprocity law. for example $(n,m)=(34,67)$. In this case one has $$\left(\frac{2}{67}\right)=(-1)^{\dfrac{67^2-1}{8}}=-1$$ and we can act as above without exhibit $\mathbb F_{67}^2$ to see that $2$ is not square.