MHB Diophantine equation has no solution

[evinda]

Hello! :)
I have a question..How can I show that the diophantine equation $15x^2-7y^2=9$ has no solution?
Could you give me a hint?

 

[I like Serena]

Re: diophantine equation has no solution

Hello! :)
I have a question..How can I show that the diophantine equation $15x^2-7y^2=9$ has no solution?
Could you give me a hint?

Hey!

Have you tried to calculate modulo a number and see if a square can fit?
Typically I would check mod 2, mod 3, mod 4, mod 5, mod 7, and mod 9.

 

[evinda]

Re: diophantine equation has no solution

Hey!

Have you tried to calculate modulo a number and see if a square can fit?
Typically I would check mod 2, mod 3, mod 4, mod 5, mod 7, and mod 9.

I thought that,as $15 x^2 \mod 3=0$ and $9 \mod 3=0$, $7y^2 \mod 3$ should also be equal to $0$.So, $3| y^2 \Rightarrow 3|y$.
So,we set $x=3^a \cdot k , y=3^b \cdot l , 3 \nmid k, 3 \nmid l, a,b \geq 0.$
Therefore,we have $3^{2a+1} \cdot 5 \cdot k^2-7 \cdot 3^{2b} \cdot l^2=3^2$
We see that it must be $2b>0 \Rightarrow b \geq 1 \Rightarrow 2b \geq 2$.
Also, $2a+1 \geq 2 \Rightarrow a \geq 1 \Rightarrow 2a+1 \geq 3$

Then we get $3^{2a-1} \cdot 5 \cdot k^2-7 \cdot 3^{2b-2} \cdot l^2=1$ with $2a-1 \geq 1$.So,it must be $2b-2=0 \Rightarrow b=1$

$3|3^{2a-1} \cdot 5 \cdot k^2=1+7 \cdot l^2$

As $3 \mid l \Rightarrow l=3n+1 \text{ or } l=3n+2$

$1+7 \cdot l^2=1+7(3m+1)=21m+8$,that is not divisible by $3$.
So,the diophantine equation $15 \cdot x^2−7 \cdot y^2=9$ has no solution.
Is it right or have I done somethig wrong? (Blush)

 

[I like Serena]

Re: diophantine equation has no solution

I thought that,as $15 x^2 \mod 3=0$ and $9 \mod 3=0$, $7y^2 \mod 3$ should also be equal to $0$.So, $3| y^2 \Rightarrow 3|y$.
So,we set $x=3^a \cdot k , y=3^b \cdot l , 3 \nmid k, 3 \nmid l, a,b \geq 0.$
Therefore,we have $3^{2a+1} \cdot 5 \cdot k^2-7 \cdot 3^{2b} \cdot l^2=3^2$
We see that it must be $2b>0 \Rightarrow b \geq 1 \Rightarrow 2b \geq 2$.
Also, $2a+1 \geq 2 \Rightarrow a \geq 1 \Rightarrow 2a+1 \geq 3$

Then we get $3^{2a-1} \cdot 5 \cdot k^2-7 \cdot 3^{2b-2} \cdot l^2=1$ with $2a-1 \geq 1$.So,it must be $2b-2=0 \Rightarrow b=1$

$3|3^{2a-1} \cdot 5 \cdot k^2=1+7 \cdot l^2$

As $3 \mid l \Rightarrow l=3n+1 \text{ or } l=3n+2$

$1+7 \cdot l^2=1+7(3m+1)=21m+8$,that is not divisible by $3$.
So,the diophantine equation $15 \cdot x^2−7 \cdot y^2=9$ has no solution.
Is it right or have I done somethig wrong? (Blush)

Looks good! (Cool)

Here's a slightly shorter version:
\begin{array}{lcrl}
15x^2−7y^2 &\equiv& 9 &\pmod 5 \\
3y^2 &\equiv& -1 &\pmod 5 \\
6y^2 &\equiv& -2 &\pmod 5 \\
y^2 &\equiv& -2 &\pmod 5
\end{array}
Since the only possibilities for $y^2$ are $0,\pm 1 \pmod 5$, there is no solution.

 

[evinda]

Re: diophantine equation has no solution

Why did you find these:

\begin{array}{lcrl}

3y^2 &\equiv& -1 &\pmod 5 \\
6y^2 &\equiv& -2 &\pmod 5 \\

\end{array}

I haven't understood it.. (Blush) Could you explain it to me? :)

 

[I like Serena]

Re: diophantine equation has no solution

Why did you find these:

I haven't understood it.. (Blush) Could you explain it to me? :)

I multiplied both sides with $2$.
That's a special number because $2$ is the inverse of $3 \pmod 5$.
That is:
$$2\cdot 3 \equiv 1 \pmod 5$$
Alternatively, we could also write it out.
It means that there is some $k \in \mathbb Z$ such that:
$$3y^2 = -1 + 5k$$
Multiply by 2 to get:
$$6y^2 = -2 + 5\cdot 2k$$
Which implies that:
$$6y^2 \equiv -2 \pmod 5$$