# Show that sys. of eq. has no integer solutions

• August 6th 2012, 01:41 AM
Show that sys. of eq. has no integer solutions
I'm to show that the system of equations below has no integer solutions.

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

Reducing to modulo 2, we get

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

This means that $x$ must be odd and the sum $x + y$ odd as well. A sum can only be odd if one number is odd and the other odd, thus $y$ 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?
• August 6th 2012, 08:48 AM
richard1234
Re: Show that sys. of eq. has no integer solutions
You may as well solve the system. Multiply first equation by -2:

$22x - 10y = 14$

$9x + 10y = -3$

Add the two equations to get $31x = 11$. Clearly, x is not an integer so we're done.
• August 6th 2012, 12:12 PM
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:

$22x - 10y = 14$

$9x + 10y = -3$

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

Yes I might as well do that but that's not the assignment.
• August 6th 2012, 01:24 PM
Soroban
Re: Show that sys. of eq. has no integer solutions

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?
• August 6th 2012, 02:09 PM
richard1234
Re: Show that sys. of eq. has no integer solutions
Quote:

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.
• August 6th 2012, 02:12 PM
richard1234
Re: Show that sys. of eq. has no integer solutions
Quote:

Originally Posted by Soroban

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 >.<
• August 8th 2012, 10:46 AM
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:

$\begin{cases} 11x - 5y \underset{2}{\equiv} 7 & \Leftrightarrow \ \ x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3 & \Leftrightarrow \ \ x \underset{2}{\equiv} 1 \end{cases}$
• August 8th 2012, 11:29 AM
richard1234
Re: Show that sys. of eq. has no integer solutions
Quote:

$\begin{cases} 11x - 5y \underset{2}{\equiv} 7 & \Leftrightarrow \ \ x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3 & \Leftrightarrow \ \ x \underset{2}{\equiv} 1 \end{cases}$
You obtain $y \equiv 0 (\mod 2)$ but that doesn't lead much to a contradiction.
$\begin{cases} 11x \equiv 7 (\mod 5) \\ 9x \equiv -3 (\mod 5) \end{cases}$
Adding, we get $20x \equiv 4 (\mod 5)$, contradiction, x is not an integer.