Show that sys. of eq. has no integer solutions

I'm to show that the system of equations below has no integer solutions.

$\displaystyle \begin{cases} 11x - 5y = 7 \: & (\text {I})\\9x + 10y = -3 \: & (\text {II}) \end{cases}$

Reducing to modulo 2, we get

$\displaystyle \begin{cases} 11x - 5y \underset{2}{\equiv} 7 & \Leftrightarrow \ \ x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3 & \Leftrightarrow \ \ x \underset{2}{\equiv} 1 \end{cases}$

This means that $\displaystyle x$ must be odd and the sum $\displaystyle x + y$ odd as well. A sum can only be odd if one number is odd and the other odd, thus $\displaystyle y$ must be an even number. I haven't seemingly arrived at a condraction, I am still however doing something wrong because the equation in fact doesn't have any integer solutions. Help anybody?

Re: Show that sys. of eq. has no integer solutions

You may as well solve the system. Multiply first equation by -2:

$\displaystyle 22x - 10y = 14$

$\displaystyle 9x + 10y = -3$

Add the two equations to get $\displaystyle 31x = 11$. Clearly, x is not an integer so we're done.

Re: Show that sys. of eq. has no integer solutions

Quote:

Originally Posted by

**richard1234** You may as well solve the system. Multiply first equation by -2:

$\displaystyle 22x - 10y = 14$

$\displaystyle 9x + 10y = -3$

Add the two equations to get $\displaystyle 31x = 11$. Clearly, x is not an integer so we're done.

Yes I might as well do that but that's not the assignment.

Re: Show that sys. of eq. has no integer solutions

Hello, MathCrusader!

Quote:

Yes, I might as well do that, but that's not the assignment.

Well, that's just great!

Could you possibly tell us: what *is* the assignment?

And do you have any more surprises for us?

Re: Show that sys. of eq. has no integer solutions

Quote:

Originally Posted by

**MathCrusader** Yes I might as well do that but that's not the assignment.

I second Soroban, what* is *the assignment? The assignment was to show that the system has no integer solution (x,y), right? It shouldn't matter *how* you do it, as long as it's valid.

Re: Show that sys. of eq. has no integer solutions

Quote:

Originally Posted by

**Soroban** Hello, MathCrusader!

And do you have any more surprises for us?

Lol...not allowed to use mods or Bezout's identity. You must use the solution I show you. I hate when teachers do that >.<

Re: Show that sys. of eq. has no integer solutions

My bad I forgot to write that it needs to be done using modular arithmetic. Considering that I was attempting to do it using modular arithmetic and I posted it under 'Number Theory', it's surprises that it wasn't evident how it's meant to be solved, but I digress.

So, how would we move on from:

$\displaystyle \begin{cases} 11x - 5y \underset{2}{\equiv} 7 & \Leftrightarrow \ \ x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3 & \Leftrightarrow \ \ x \underset{2}{\equiv} 1 \end{cases}$

Re: Show that sys. of eq. has no integer solutions

Quote:

Originally Posted by

**MathCrusader** My bad I forgot to write that it needs to be done using modular arithmetic. Considering that I was attempting to do it using modular arithmetic and I posted it under 'Number Theory', it's surprises that it wasn't evident how it's meant to be solved, but I digress.

So, how would we move on from:

$\displaystyle \begin{cases} 11x - 5y \underset{2}{\equiv} 7 & \Leftrightarrow \ \ x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3 & \Leftrightarrow \ \ x \underset{2}{\equiv} 1 \end{cases}$

You obtain $\displaystyle y \equiv 0 (\mod 2)$ but that doesn't lead much to a contradiction.

If we look at it mod 5, we obtain

$\displaystyle \begin{cases} 11x \equiv 7 (\mod 5) \\ 9x \equiv -3 (\mod 5) \end{cases}$

Adding, we get $\displaystyle 20x \equiv 4 (\mod 5)$, contradiction, x is not an integer.