# Prove if $x^{2}-5xy-3$ is even, then $x+y$ is odd, where $x,y \in\mathbb{Z}$

I know for you this is easy but for me is not. I give my best shot but it's no use so I need someone to teach about all this stuff.

As I try to solve this one, I come up with this answer:

Suppose $x^2-5xy-3$ is even, then $x=2a + 1$ and $y=2b$ for some integers $a,b \in\mathbb{Z}$.

Thus, \begin{aligned}x^2-5xy-3&=(2a+1)^2-5(2a+1)(2b)-3 \\ &=(4a^2+4a+1)-20ab+10b-3 \\ &=2(2a^2+2a)-20ab+10b-2 \\ &=\;?\end{aligned}

And I don't know what's the next step. I know there's something wrong with my procedure.

I also have plenty of other questions that need to be answered. I've already answer this equations but I can't solve it.

• If $m$ is odd and $n$ is even, then $m^2-5mn+n^2+1$ is even, where $m,n\in\mathbb{Z}$.
• If $x-y$ is even, then $x^2+3xy-5$ is odd, where $m,n \in\mathbb{Z}$.
• Let $a,b \in\mathbb{Z}$. If $2b^2-3ab+1$ is even, then $2a-b$ is odd.
• Let $m,n \in\mathbb{Z}$. Prove that if $m^2+1$ is even, then $2n+m$ is even.

Even though I'm not good in math, I know in the future I will be good in math by practicing and with your help.

This not a assignment, I'm practicing solving problems like this to be good in math.

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If x^2 - 5xy - 3 is even, then x^2 - 5xy is odd. But then any integral factor of this expression must be odd, too. –  franz lemmermeyer Jul 15 '12 at 13:35

Here is how to do the first problem you mentioned within the question (the one where you showed your work) - it should give you the general idea.

Key fact A: An integer $t$ is even if and only if $t=2k$ for some integer $k$; and similarly, $t$ is odd if and only if $t=2k+1$ for some integer $k$.

Key fact B: $$\begin{array}{c|c|c|} + & \bf\text{odd} & \bf\text{even}\\\hline \bf\text{odd} & \text{even} & \text{odd}\\\hline \bf\text{even} & \text{odd} & \text{even}\\\hline \end{array}$$

Key fact C: $$\begin{array}{c|c|c|} \times & \bf\text{odd} & \bf\text{even}\\\hline \bf\text{odd} & \text{odd} & \text{even}\\\hline \bf\text{even} & \text{even} & \text{even}\\\hline \end{array}$$

Now suppose that, for some $x,y\in\mathbb{Z}$, the quantity $$x^2-5xy-3=(x^2-5xy)+(-3)$$ is even. Because $-3$ is odd, this is only possible if $x^2-5xy$ is odd (look at key fact B).

We can factor $x^2-5xy$ as $(x)\times(x-5y)$. The only way that $x^2-5xy$ can be odd is if both $x$ and $x-5y$ are odd (look at key fact C).

Because $x$ is odd and $x-5y=(x)+(-5y)$ is odd, we can see that $-5y$ must be even (key fact B).

Because $-5y=(-5)\times (y)$ is even and $-5$ is odd, it must be the case that $y$ is even (key fact C).

Thus, we have shown, starting from the knowledge that $x^2-5xy-3$ is odd, that $x$ must be odd and $y$ must be even. By key fact A, there must be an integer $a$ such that $x=2a+1$, and there must be an integer $b$ such that $y=2b$.

Lastly, using key fact B, the fact that $x$ is odd and $y$ is even implies that $x+y$ is odd.

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HINT: Note that $x(x-5y)$ is odd and any odd number can have only odd factors. Hence, $x$ and $x-5y$ are both odd. Can you now conclude that $x+y$ is odd by showing that $y$ is even?

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by showing that $y$ is even... Or, by writing $(x+y)=(x-5y)+6y$, hence $x+y$ and $x-5y$ have the same parity. –  Did Jul 15 '12 at 14:07

Below are a sketches of a few proofs. Take your pick. The first proof uses simple parity arithmetic (LHS tables below), the rest use mod $2$ arithmetic (RHS tables), where $\rm\:even\leftrightarrow 0,\,\ odd \leftrightarrow 1$

$$\begin{array}{rcl} \rm Parity\ Arithmetic & &\rm\!\!\! modulo\ 2 \\ \begin{array}{|c|c|c|}\hline + & \bf\text{even} & \bf\color{#0A0}{odd}\\\hline \bf\text{even} & \text{even} & \text{odd}\\\hline \bf\color{blue}{odd} & \text{odd} &\rm \bf\color{#C00}{even}\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline + & \bf 0 & \bf 1\\\hline \bf 0 & 0 & 1\\\hline \bf 1 & 1 & 0\\\hline \end{array} \\ \begin{array}{|c|c|c|}\hline \times & \bf\text{even} & \bf\color{#0AA}{odd}\\\hline \bf\text{even} & \text{even} & \text{even}\\\hline \bf\color{#0AA}{odd} & \text{even} & \bf\color{blue}{odd}\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline \times & \bf 0 & \bf\color{#0AA}1 \\\hline \bf 0 & 0 & 0\\\hline \bf\color{#0AA} 1 & 0 & \bf\color{blue}1\\\hline \end{array} \end{array}$$

$\rm (1)\ \ \ x\,(x\!+\!y)\, - 6xy\!-\!3\ \ \bf\color{#C00}{even},$ $\rm\, -6xy\!-\!3\ \ \bf\color{#0A0}{odd}$ $\:\Rightarrow$ $\rm\: x\,(x\!+\!y)\ \ \bf\color{blue}{odd}$ $\rm\:\Rightarrow\: x\!+\!y\ \ \bf\color{#0AA}{odd}$

$\rm(2)\ \ \ mod\ 2\!:\ \ 0 \,\equiv\, x^2\!-5xy-3 \,\equiv\, x^2\!+xy-1\:\Rightarrow\: x(x\!+\!y)\,\equiv \,{\bf\color{blue}1}\:\Rightarrow\:x\!+\!y\,\equiv\, {\bf\color{#0AA}1}$

$\rm(3)\ \$ If not, $\rm\:x\!+\!y\equiv 0\:\Rightarrow\:x \equiv -y\,$ $\Rightarrow$ $\rm\:x^2\!-\!5xy\!-\!3\equiv 6y^2\!-\!3\equiv 1,\:$ contra hypothesis.

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There are 4 cases:EE $x$ even, $y$ even; EO $x$ even, $y$ odd; OE $x$ odd $y$ even; OO $x$ odd, $y$ odd.

In the first two cases $x^2-5xy=x(x-5y)$ is even so $x^2-5xy-3$ is odd. So our $x, y$ can only occur for the last two cases. In the last case OO: $x(x-5y)$ is even since $x-5y$ is even and therefore $x^2-5xy-3$ is odd.

Thus the only possibility is case OE. In this case $x+y$ is odd.

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Perhaps something slightly different and shorter: supppose $\,x,y\,$ have the same parity, then $\,x(x-5y)\,$ is always even (if they both are even then this is the product of even integers, and if they both are odd then $\,5y\,$ is odd and thus $\,x-5y\,$ is even), so $\,x(x-5y)-3=x^2-5xy-3\,$ cannot be even...contradiction!

Thus, $\,x,y\,$ have different parity $\,\Longleftrightarrow x+y\,$ is odd.

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Nothing new and still,

given an integer $a$ which is a polynomial expression of two integers $x,y$ with integer coefficients (an expression like your $x^2-5xy-3$ which is a sum of products of $x,y$ and integer numbers), the parities of $x$ and $y$ determine the parity of $a$ (using key facts B and C from Zev's answer). Thus, any statement about the connection between the parities of the above numbers can be checked by going over all the possible cases. Of course, you can employ some shortcuts like the ones suggested in the answers, but this is the ultimate "key fact" and it can be broadly generalized (more variables, different modulus and so on).

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