No integer Solutions

**Learner248** · July 14th 2011, 02:51 AM

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

**Drexel28** · July 14th 2011, 11:53 AM

Originally Posted by Learner248

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 $x,y$ were solutions then $7y^2=9-15x^2\equiv 0\text{ mod }3$ and so $3\mid \y^2$ but since $3$ is prime and $3\nmid 7$ we may conclude that $3\mid y$ and so $y=3y'$ and so we must have that $15x^2-63y'^2=9$ but in the same spirit we see that this implies $9\mid 15x^2$ and so $3\mid 5x^2$ and since $3$ is prime and $3\nmid$ we may conclude that $3\mid x$ and so $x=3x'$ and so our equation becomes, after dividing both sides by $9$ , the following $15x'^2-7y'^2=1$ but this says that $y'^2\equiv 7y'^2\equiv -1\equiv 2\text{ mod }3$ but it's evident that $2$ isn't a quadratic residue $\text{ mod }3$ /

**Soroban** · July 14th 2011, 02:50 PM

Hello, Learner248!

I am puzzled, too.

Prove that: $15x^2 -7y^2\:=\:9$ has no integer solutions.

I have understood a few steps

$15x^2 -7y^2\:=\:9$

$y=3y_1 \quad\Rightarrow\quad 5x^2-21y^2\:=\:3$

$x-3x_1\quad\Rightarrow\quad 15x_1^2-7y_1^2\:=\:1$

$y_1^2\:=\:2\text{ (mod 3)}$

But this contradiction how?

If there is any other way, plz tell.

We have: . $15x^2 - 7y^2 \:=\:9$

Since $15x^2$ and $9$ are multiples of 3,
. . then $7y^2$ must be a multiple of 3.
Hence, $y$ is a multiple of 3: . $y \:=\:3y_1$

The equation becomes: . $15x^2 - 7(3y_1)^2 \:=\:9 \quad\Rightarrow\quad 15x^2 - 63y_1^2 \:=\:9$
. . Divide by 3: . $5x^2 - 21y_1^2 \:=\:3$

Since $21y_1^2$ and $3$ are multiples of 3,
. . then $5x^2$ must be a multiple of 3.
Hence, $x$ is a multiple of 3: . $x \:=\:3x_1$

The equation becomes: . $5(3x_1)^2 - 21y_1^2 \:=\:3 \quad\Rightarrow\quad 45x_1^2 - 21y_1^2 \:=\:3$
. . Divide by 3: . $15x_1^2 - 7y_1^2 \:=\:1$

And I have no idea where they got: $y_1^2 = 2\text{ (mod 3)}$
. . It doesn't seem to be true.

I think I can finish the proof . . .

We have: . $15x_1^2 - 7y_1^2 \:=\:1$

Then: . $y_1^2 \:=\:\frac{15x_1^2-1}{7} \quad\Rightarrow\quad y_1^2 \:=\:2x_1^2 + \frac{x_1^2-1}{7}$ .[1]

Since $y_1^2$ is an integer, then $x_1^2-1$ must be a multiple of 7.
. . That is: . $x_1^2-1 \:=\:7k \quad\Rightarrow\quad x_1^2 \:=\:7k+1$

Substitute into [1]: . $y_1^2 \:=\:2(7k+1) + \frac{(7k+1)-1}{7}$
. . which simplfies to: . $y_1^2 \:=\:15k+2$

For integer values of $k,\:y_1^2$ will end in 2 or 7.

But squares must end in $\{0,1,4,5,6,9\}$

There is our contradiction!

**Opalg** · July 14th 2011, 11:34 PM

Originally Posted by Learner248

Prove that $15x^2 -7y^2=9$ has no integer solutions.

I have understood a few steps

$15x^2 -7y^2=9 \;\Rightarrow\; \ldots \;\Rightarrow\; y_1^2=2 \pmod 3$

But this contradiction how???

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

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

So the equation $y_1^2=2 \pmod 3$ has no solutions.

**Learner248** · July 16th 2011, 03:01 AM

Thanks Everybody

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

Anyways

Thanx a lot

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