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

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- Jul 14th 2011, 02:51 AMLearner248No integer Solutions
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 - Jul 14th 2011, 11:53 AMDrexel28Re: No integer Solutions
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 $\displaystyle \text{ mod }3$/

- Jul 14th 2011, 02:50 PMSorobanRe: No integer Solutions
Hello, Learner248!

I am puzzled, too.

Quote:

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 $\displaystyle \{0,1,4,5,6,9\}$

is our contradiction!*There*

- Jul 14th 2011, 11:34 PMOpalgRe: No integer Solutions
There are no numbers with square congruent to 2 mod 3. In fact, every integer is congruent to 0, 1 or 2 (mod 3), and

$\displaystyle 0^2\equiv0\pmod3,\quad 1^2\equiv1\pmod3,\quad 2^2\equiv1\pmod3.$

So the equation $\displaystyle y_1^2=2 \pmod 3$ has no solutions. - Jul 16th 2011, 03:01 AMLearner248Re: No integer Solutions
Thanks Everybody

I figured it out a few minutes later after I posted it.

Anyways

Thanx a lot(Yes)