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...
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?
Hello, kohila!
I have a primitive solution with brute-force Listing.
$\displaystyle \text{If }a,b,c,d\text{ are four positive integers such that }a+b+c\text{ is even}$
$\displaystyle \text{and }b+c+d\text{ odd, then }a^2 - d^2\text{ is neccesarily }\_\_\_.$
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.
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.