Show that sys. of eq. has no integer solutions

**MathCrusader** · August 6th 2012, 01:41 AM

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?

**richard1234** · August 6th 2012, 08:48 AM

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.

**MathCrusader** · August 6th 2012, 12:12 PM

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.

**Soroban** · August 6th 2012, 01:24 PM

Hello, MathCrusader!

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?

**richard1234** · August 6th 2012, 02:09 PM

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.

**richard1234** · August 6th 2012, 02:12 PM

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 >.<

**MathCrusader** · August 8th 2012, 10:46 AM

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:

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

**richard1234** · August 8th 2012, 11:29 AM

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:

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

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

If we look at it mod 5, we obtain

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

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

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