MHB Diophantine equation has no solution

  • Thread starter Thread starter evinda
  • Start date Start date
AI Thread Summary
The discussion centers on proving that the Diophantine equation 15x² - 7y² = 9 has no solutions. Participants suggest using modular arithmetic, specifically checking the equation modulo 3 and 5. Through calculations, it is shown that y must be divisible by 3, leading to a contradiction when substituting back into the equation. Ultimately, the conclusion is reached that the equation has no solution, supported by the modular analysis. The participants confirm the correctness of the reasoning presented.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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:
 
Mathematics news on Phys.org
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! :o

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.
 
Re: diophantine equation has no solution

I like Serena said:
Hey! :o

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)
 
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.
 
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? :)
 
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:
$$6y^2 \equiv -2 \pmod 5$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top