Question

Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

• A. The minimum possible value of \(a^2 + b^2 + c^2 + d^2\) is \(4m^2 - 2m + 1\)
• B. The minimum possible value of \(a^2 + b^2 + c^2 + d^2\) is \(4m^2 + 2m + 1\)
• C. The maximum possible value of \(a^2 + b^2 + c^2 + d^2\) is \(4m^2 - 2m + 1\)
• D. The maximum possible value of \(a^2 + b^2 + c^2 + d^2\) is \(4m^2 + 2m + 1\)

Hint:

For a fixed total, the sum of squares is smallest when the four integers are as close to each other as possible. Ask also whether the sum of squares can ever stop growing.

Answer 1

The Correct Option is B

Step 1: Understand what is fixed and what changes:
The four numbers are integers and their sum is fixed at \(4m+1\). We are asked about the sum of their squares \(a^2+b^2+c^2+d^2\), which changes as the numbers change.
Two of the options talk about a minimum and two about a maximum, so first we must decide whether a minimum or a maximum can exist at all.

Step 2: Write each number as m plus a small integer:
The average of the four numbers is \(m + \tfrac{1}{4}\), so it is natural to write \(a = m + e_1\), \(b = m + e_2\), \(c = m + e_3\), \(d = m + e_4\), where each \(e_i\) is an integer.
The sum condition gives \(4m + (e_1+e_2+e_3+e_4) = 4m + 1\), so \(e_1+e_2+e_3+e_4 = 1\).

Step 3: Expand the sum of squares:
\[ a^2+b^2+c^2+d^2 = \sum (m+e_i)^2 = 4m^2 + 2m\sum e_i + \sum e_i^2 \]
Since \(\sum e_i = 1\), this becomes
\[ a^2+b^2+c^2+d^2 = 4m^2 + 2m + (e_1^2+e_2^2+e_3^2+e_4^2) \]
So everything depends on the quantity \(T = e_1^2+e_2^2+e_3^2+e_4^2\), where the \(e_i\) are integers adding up to 1.

Step 4: Find the smallest value of T:
Each \(e_i\) is an integer, so \(e_i^2 \geq |e_i|\). Therefore \(T \geq |e_1|+|e_2|+|e_3|+|e_4| \geq |e_1+e_2+e_3+e_4| = 1\).
T equals 1 when one of the \(e_i\) is 1 and the other three are 0. That is, the numbers are \(m, m, m, m+1\). These add up to \(4m+1\), as required.
So the smallest possible sum of squares is \(4m^2 + 2m + 1\), and this value is actually reached.

Step 5: Show that no maximum exists:
Take the numbers \(N, -N, 2m, 2m+1\) for any large integer N. They add up to \(4m+1\), and their squares add up to \(2N^2 + 4m^2 + (2m+1)^2\).
This grows without limit as N grows, so there is no maximum value at all.

Step 6: Check each option:
Option (1) says the minimum is \(4m^2 - 2m + 1\). This is smaller than the true minimum \(4m^2+2m+1\) for every positive m, so it can never be reached. FALSE.
Option (2) says the minimum is \(4m^2 + 2m + 1\). This matches our result. TRUE.
Options (3) and (4) talk about a maximum, but no maximum exists. FALSE.

Final Answer:
The only statement that must be true is that the minimum possible value of the sum of squares is \(4m^2+2m+1\), which is option (2). \[ \boxed{\text{Option (2)}} \]