Prove that 15x^2 -7y^2=9 has no integer solutions.
I have understood a few steps
15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3
But this contradiction how???
If there is any other way plz tell.
Thanx
Prove that 15x^2 -7y^2=9 has no integer solutions.
I have understood a few steps
15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3
But this contradiction how???
If there is any other way plz tell.
Thanx
I'm sorry that's a little hard to make out, presumably what it says is that if $\displaystyle x,y$ were solutions then $\displaystyle 7y^2=9-15x^2\equiv 0\text{ mod }3$ and so $\displaystyle 3\mid \y^2$ but since $\displaystyle 3$ is prime and $\displaystyle 3\nmid 7$ we may conclude that $\displaystyle 3\mid y$ and so $\displaystyle y=3y'$ and so we must have that $\displaystyle 15x^2-63y'^2=9$ but in the same spirit we see that this implies $\displaystyle 9\mid 15x^2$ and so $\displaystyle 3\mid 5x^2$ and since $\displaystyle 3$ is prime and $\displaystyle 3\nmid$ we may conclude that $\displaystyle 3\mid x$ and so $\displaystyle x=3x'$ and so our equation becomes, after dividing both sides by $\displaystyle 9$, the following $\displaystyle 15x'^2-7y'^2=1$ but this says that $\displaystyle y'^2\equiv 7y'^2\equiv -1\equiv 2\text{ mod }3$ but it's evident that $\displaystyle 2$ isn't a quadratic residue $\displaystyle \text{ mod }3$/
Hello, Learner248!
I am puzzled, too.
Prove that: $\displaystyle 15x^2 -7y^2\:=\:9$ has no integer solutions.
I have understood a few steps
$\displaystyle 15x^2 -7y^2\:=\:9$
$\displaystyle y=3y_1 \quad\Rightarrow\quad 5x^2-21y^2\:=\:3$
$\displaystyle x-3x_1\quad\Rightarrow\quad 15x_1^2-7y_1^2\:=\:1$
$\displaystyle y_1^2\:=\:2\text{ (mod 3)}$
But this contradiction how?
If there is any other way, plz tell.
We have: .$\displaystyle 15x^2 - 7y^2 \:=\:9$
Since $\displaystyle 15x^2$ and $\displaystyle 9$ are multiples of 3,
. . then $\displaystyle 7y^2$ must be a multiple of 3.
Hence, $\displaystyle y$ is a multiple of 3: .$\displaystyle y \:=\:3y_1$
The equation becomes: .$\displaystyle 15x^2 - 7(3y_1)^2 \:=\:9 \quad\Rightarrow\quad 15x^2 - 63y_1^2 \:=\:9$
. . Divide by 3: .$\displaystyle 5x^2 - 21y_1^2 \:=\:3$
Since $\displaystyle 21y_1^2$ and $\displaystyle 3$ are multiples of 3,
. . then $\displaystyle 5x^2$ must be a multiple of 3.
Hence, $\displaystyle x$ is a multiple of 3: .$\displaystyle x \:=\:3x_1$
The equation becomes: .$\displaystyle 5(3x_1)^2 - 21y_1^2 \:=\:3 \quad\Rightarrow\quad 45x_1^2 - 21y_1^2 \:=\:3$
. . Divide by 3: .$\displaystyle 15x_1^2 - 7y_1^2 \:=\:1$
And I have no idea where they got: $\displaystyle y_1^2 = 2\text{ (mod 3)}$
. . It doesn't seem to be true.
I think I can finish the proof . . .
We have: .$\displaystyle 15x_1^2 - 7y_1^2 \:=\:1$
Then: .$\displaystyle y_1^2 \:=\:\frac{15x_1^2-1}{7} \quad\Rightarrow\quad y_1^2 \:=\:2x_1^2 + \frac{x_1^2-1}{7}$ .[1]
Since $\displaystyle y_1^2$ is an integer, then $\displaystyle x_1^2-1$ must be a multiple of 7.
. . That is: .$\displaystyle x_1^2-1 \:=\:7k \quad\Rightarrow\quad x_1^2 \:=\:7k+1$
Substitute into [1]: .$\displaystyle y_1^2 \:=\:2(7k+1) + \frac{(7k+1)-1}{7} $
. . which simplfies to: .$\displaystyle y_1^2 \:=\:15k+2$
For integer values of $\displaystyle k,\:y_1^2$ will end in 2 or 7.
But squares must end in $\displaystyle \{0,1,4,5,6,9\}$
There is our contradiction!