Diophantine equation has no solution

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In summary, we can show that the diophantine equation $15x^2-7y^2=9$ has no solution by checking the equation modulo different numbers and finding that there is no possible value for $y^2$ that satisfies the equation. Another way to look at it is by using the fact that $2$ is the inverse of $3 \pmod 5$, allowing us to manipulate the equation in a more convenient form.
  • #1
evinda
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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? :rolleyes:
 
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  • #2
Re: diophantine equation has no solution

evinda said:
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? :rolleyes:

Hey! :eek:

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.
 
  • #3
Re: diophantine equation has no solution

I like Serena said:
Hey! :eek:

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)
 
  • #4
Re: diophantine equation has no solution

evinda said:
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.
 
  • #5
Re: diophantine equation has no solution

Why did you find these:

I like Serena said:
\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? :)
 
  • #6
Re: diophantine equation has no solution

evinda said:
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:
$$6y^2 \equiv -2 \pmod 5$$
 

FAQ: Diophantine equation has no solution

What is a Diophantine equation?

A Diophantine equation is an algebraic equation where the solutions are required to be whole numbers (integers). It is named after the ancient Greek mathematician Diophantus.

How do you know if a Diophantine equation has no solution?

A Diophantine equation has no solution if it is impossible to find whole number values for all of the variables that make the equation true. This can be determined through various mathematical techniques, such as using modular arithmetic or considering the prime factorization of the coefficients.

Why is it important to study Diophantine equations with no solutions?

Studying Diophantine equations with no solutions can help us understand the limitations and structure of number systems. It also has applications in cryptography and coding theory.

Are there any famous examples of Diophantine equations with no solutions?

Yes, one famous example is Fermat's Last Theorem which states that there are no whole number solutions to the equation an + bn = cn for any integer value of n greater than 2. This theorem was proven by Andrew Wiles in 1994 after over 350 years of attempts by mathematicians.

Is there a general method for solving Diophantine equations with no solutions?

No, there is no general method for solving Diophantine equations with no solutions. However, there are specific techniques and strategies that can be used depending on the form of the equation. In some cases, it may also be helpful to use computer algorithms to find solutions or prove that none exist.

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