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

Please explain clearly...(Crying)

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- Oct 22nd 2012, 05:10 AMkohilaEven 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

Please explain clearly...(Crying) - Oct 22nd 2012, 06:56 AMPlatoRe: Even numbers and odd numbers..
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 $\displaystyle a-d=2K-1$ which is odd and $\displaystyle a+d=2J+1-2b-2c$ which is also odd.

But $\displaystyle a^2-b^2=(a-b)(a+b)$ SO? - Oct 22nd 2012, 04:07 PMSorobanRe: Even numbers and odd numbers..
Hello, kohila!

I have a primitive solution with brute-force Listing.

Quote:

$\displaystyle \text{If }a,b,c,d\text{ are four positive integers such that }a+b+c\text{ is even}$

$\displaystyle \text{and }b+c+d\text{ odd, then }a^2 - d^2\text{ is neccesarily }\_\_\_.$

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

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

Hence, there are four cases to consider:

. . (1) $\displaystyle a,b,c$ are even.

. . (2) $\displaystyle a$ even, $\displaystyle b,c$ odd.

. . (3) $\displaystyle b$ even, $\displaystyle a,c$ odd.

. . (4) $\displaystyle c$ even, $\displaystyle a,b$ odd.

$\displaystyle (1)\;[a\;b\;c\;d] \,=\,[e\;e\;e\;\_\,]$

. . .Since $\displaystyle b+c+d$ is odd,$\displaystyle d$ must be odd.

. . .Hence: $\displaystyle a$ is even, $\displaystyle d$ is odd.

$\displaystyle (2)\;[a\;b\;c\;d] \,=\,[e\;o\;o\;\_\,]$

. . .Since $\displaystyle b+c+d$ is odd, $\displaystyle d$ must be odd.

. . .Hence, $\displaystyle a$ is even, $\displaystyle d$ is odd.

$\displaystyle (3)\;[a\;b\;c\;d] \,=\,[o\;e\;o\;\_\,]$

. . .Since $\displaystyle b+c+d$ is odd, $\displaystyle d$ must be even.

. . .Hence, $\displaystyle a$ is odd, $\displaystyle d$ is even.

$\displaystyle (4)\;[a\;b\;c\;d] \,=\,[o\;o\;e\;\_\,]$

. . .Since $\displaystyle b+c+d$ is odd, $\displaystyle d$ must be even.

. . .Hence, $\displaystyle a$ is odd, $\displaystyle d$ is even.

In all cases, $\displaystyle a$ and $\displaystyle d$ have opposite parity.

Hence, their squares will have opposite parity.

The difference between two numbers of opposite parity is.*odd*

- Oct 22nd 2012, 06:56 PMrichard1234Re: Even numbers and odd numbers..
Necessarily...odd? Also, a,b,c,d should be integers, otherwise we cannot claim anything.

We have

$\displaystyle a + b + c \equiv 0 (\mod 2)$

$\displaystyle b + c + d \equiv 1 (\mod 2)$

If we subtract the second equation from the first, we get $\displaystyle a-d \equiv 1 (\mod 2)$, therefore a-d is odd. This implies that a+d is also odd, since a+d = (a-d) + 2d. Therefore $\displaystyle (a-d)(a+d) = a^2 - d^2$ must be odd. - Oct 22nd 2012, 07:31 PMkohilaRe: Even numbers and odd numbers..
**Thanks... I can get it from your explanation.** - Oct 22nd 2012, 07:45 PMkohilaRe: Even numbers and odd numbers..
**Much thanks for ur deep core conceptual clarification...**