# Thread: Even numbers and odd numbers..

1. ## Even numbers and odd numbers..

If a,b,c and d are four positive real numbers such that the sum of a,b and c is even and the sum of b,c and d is odd, then a^2 - d^2 is neccesarily

2. ## Re: Even numbers and odd numbers..

Originally Posted by kohila
If a,b,c and d are four positive real numbers such that the sum of a,b and c is even and the sum of b,c and d is odd, then a^2 - d^2 is neccesarily
I think that you mean "a,b,c and d are four positive integers" otherwise no conclusion is possible.
If that is the case then you can show $a-d=2K-1$ which is odd and $a+d=2J+1-2b-2c$ which is also odd.

But $a^2-b^2=(a-b)(a+b)$ SO?

3. ## Re: Even numbers and odd numbers..

Hello, kohila!

I have a primitive solution with brute-force Listing.

$\text{If }a,b,c,d\text{ are four positive integers such that }a+b+c\text{ is even}$
$\text{and }b+c+d\text{ odd, then }a^2 - d^2\text{ is neccesarily }\_\_\_.$

Let $\begin{Bmatrix} e &=& \text{even} \\ o &=& \text{odd}\end{Bmatrix}$

Since $a+b+c$ is even, either (1) all are even or (2) exactly one is even.

Hence, there are four cases to consider:
. . (1) $a,b,c$ are even.
. . (2) $a$ even, $b,c$ odd.
. . (3) $b$ even, $a,c$ odd.
. . (4) $c$ even, $a,b$ odd.

$(1)\;[a\;b\;c\;d] \,=\,[e\;e\;e\;\_\,]$
. . .Since $b+c+d$ is odd, $d$ must be odd.
. . .Hence: $a$ is even, $d$ is odd.

$(2)\;[a\;b\;c\;d] \,=\,[e\;o\;o\;\_\,]$
. . .Since $b+c+d$ is odd, $d$ must be odd.
. . .Hence, $a$ is even, $d$ is odd.

$(3)\;[a\;b\;c\;d] \,=\,[o\;e\;o\;\_\,]$
. . .Since $b+c+d$ is odd, $d$ must be even.
. . .Hence, $a$ is odd, $d$ is even.

$(4)\;[a\;b\;c\;d] \,=\,[o\;o\;e\;\_\,]$
. . .Since $b+c+d$ is odd, $d$ must be even.
. . .Hence, $a$ is odd, $d$ is even.

In all cases, $a$ and $d$ have opposite parity.
Hence, their squares will have opposite parity.

The difference between two numbers of opposite parity is odd.

4. ## Re: Even numbers and odd numbers..

Originally Posted by kohila
If a,b,c and d are four positive real numbers such that the sum of a,b and c is even and the sum of b,c and d is odd, then a^2 - d^2 is neccesarily

Necessarily...odd? Also, a,b,c,d should be integers, otherwise we cannot claim anything.

We have
$a + b + c \equiv 0 (\mod 2)$
$b + c + d \equiv 1 (\mod 2)$

If we subtract the second equation from the first, we get $a-d \equiv 1 (\mod 2)$, therefore a-d is odd. This implies that a+d is also odd, since a+d = (a-d) + 2d. Therefore $(a-d)(a+d) = a^2 - d^2$ must be odd.

5. ## Re: Even numbers and odd numbers..

Thanks... I can get it from your explanation.

6. ## Re: Even numbers and odd numbers..

Much thanks for ur deep core conceptual clarification...