Question

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:

Hint:

For minimizing sums of squares with a fixed sum constraint, keep the numbers as close together as possible. Here, that meant taking $a,b,c,d$ near the average $\frac{46}{4}=11.5$, and distributing small integer deviations evenly.

Answer 1

Correct Answer: 2

Approach: Three squared gaps are smallest when \(a\) sits right in the middle of \(b, c, d\). Since the four integers must sum to 46 and 46 is not divisible by 4, we cannot make all four equal, so we bunch them as tightly as possible around \(a\).

Step 1: What we are minimising.
We want \((a-b)^2 + (a-c)^2 + (a-d)^2\) as small as possible. Each square is \(0\) only when that number equals \(a\). So the dream case is \(a=b=c=d\).

Step 2: Test the dream case.
\(a=b=c=d\) forces \(4a = 46\), i.e. \(a = 11.5\), which is not an integer. So at least one of \(b, c, d\) must differ from \(a\).

Step 3: Spread the leftover by exactly 1.
Keep the integers as close to \(11.5\) as possible: take two of them \(=11\) and two \(=12\). For example \(a = 12,\ b = 12,\ c = 11,\ d = 11\).
Check sum: \(12 + 12 + 11 + 11 = 46\). Good.

Step 4: Evaluate.
\[ (12-12)^2 + (12-11)^2 + (12-11)^2 = 0 + 1 + 1 = 2. \]
We cannot do better: only one of \(b, c, d\) can equal \(a\), so at least two squares are \(\ge 1\), forcing the sum \(\ge 2\).

Minimum value = 2.

Answer 2

Reframing the problem. Write \(b=a+x,\ c=a+y,\ d=a+z\) for integers \(x,y,z\). Since \(a+b+c+d=46\), we get \(4a+x+y+z=46\), so \(x+y+z=46-4a\). The quantity to minimize, \((a-b)^2+(a-c)^2+(a-d)^2\), equals \(x^2+y^2+z^2\).

For a fixed integer sum \(S=x+y+z\), the sum of squares is smallest when \(x,y,z\) are as close to each other as possible. Trying the two integers nearest to \(46/4=11.5\):
For \(a=11\): \(x+y+z=2\), most even split \(\{1,1,0\}\), giving \(1^2+1^2+0^2=2\).
For \(a=12\): \(x+y+z=-2\), most even split \(\{-1,-1,0\}\), again giving \(2\).
Any other value of \(a\) pushes \(x+y+z\) further from 0 and only increases the minimum achievable sum of squares.

So the minimum value is \(2\), achieved for instance at \(a=11,\ b=12,\ c=12,\ d=11\) (sum \(=46\)).

Answer: 2.