If a,b,c and d are integers such that their sum is 46, then the minimum possible value of    (a − b)² + (a −c)² + (a − d)² is

The correct answer is 2.

Let's break down the given information step-by-step. We are told that we have four integers: $a$, $b$, $c$, and $d$. We are also given their sum, which allows us to write our first equation:

$a+b+c+d=46$

Our objective is to find the minimum possible value of the following expression:

${(a-b)}^2+{(a-c)}^2+{(a-d)}^2$

To make solving this easier, let's introduce variables for the differences being squared. These represent how far away $b$, $c$, and $d$ are from $a$. Let:

$x=a-b$
$y=a-c$
$z=a-d$

By rearranging these new definitions, we can express $b$, $c$, and $d$ in terms of $a$:

$b=a-x$
$c=a-y$
$d=a-z$

Now, let's substitute these expressions back into our original sum equation:

$a+(a-x)+(a-y)+(a-z)=46$

Combining the $a$ terms yields:

$4a-(x+y+z)=46$

Let's isolate the sum of our new variables, $x+y+z$, by rearranging the equation:

$x+y+z=4a-46$

We can factor a 2 out of the right side to reveal an important property:

$x+y+z=2(2a-23)$

Since the problem states that $a$ is an integer, we know that $2a-23$ must also be an integer. When you multiply any integer by 2, the result is an even number. Therefore, the sum $x+y+z$ must be an even integer.

Returning to our objective, we want to find the minimum possible value of:

$x^2+y^2+z^2$

To minimize a sum of squares of integers, the integers themselves need to be as close to 0 as possible. Let's systematically test the smallest possible non-negative integer values for the sum of squares.

Could the minimum be 0?
For the sum of squares to equal 0, each individual squared term must be 0. This means $x=0$, $y=0$, and $z=0$. This would make their sum $x+y+z=0$.
If we substitute this into our formula $4a-(x+y+z)=46$, we get $4a-0=46$, which solves to $a=11.5$. Because $a$ must be an integer, this scenario is impossible.

Could the minimum be 1?
For the sum of three squares to equal 1, the values of $(x,y,z)$ must be some combination of $\pm 1$, 0, and 0 (for example: 1, 0, 0 or -1, 0, 0).
In any of these combinations, the sum $x+y+z$ would equal either 1 or -1. However, we proved earlier that $x+y+z$ must be an even integer. Because 1 and -1 are odd, this scenario is also impossible.

Could the minimum be 2?
For the sum of three squares to equal 2, the values of $(x,y,z)$ must be some combination of two $\pm 1$s and one 0 (for example: 1, 1, 0 or -1, 1, 0).
Let's see if we can get an even sum from these. If we pick $x=1$, $y=1$, and $z=0$, their sum is $1+1+0=2$. This is an even number! Let's check if it produces an integer for $a$:

$4a-\left(2\right)=46$
$4a=48$
$a=12$

Since $a=12$ is indeed an integer, this is a perfectly valid solution. We can verify the rest of the original numbers just to be sure:

$b=12-1=11$
$c=12-1=11$
$d=12-0=12$

Checking the sum: $12+11+11+12=46$. And finally, the value of the expression is $1^2+1^2+0^2=2$. Because 0 and 1 were impossible, 2 is confirmed to be the absolute minimum possible value.