Problem 1

Given any set

A=\{a_1,a_2,a_3,a_4\}

of four distinct positive integers, we denote the sum

a_1+a_2+a_3+a_4

s_A

. Let

n_A

denote the number of pairs

(i,j)

with

1\le i<j\le4

for which

a_i+a_j

divides

s_A

. Find all sets

A

of four distinct positive integers which achieve the largest possible value of

n_A

Step 5 of 5: Solve both cases and verify

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In plain words

Each admissible (n,m) pins down all four numbers as fixed fractions of sA, and both resulting patterns really do achieve nA=4.

A=\{k,5k,7k,11k\}\ \text{or}\ A=\{k,11k,19k,29k\}

Detailed analysis

For $(n,m)=(3,4)$ , solving the linear system gives $a_1=\dfrac{s_A}{24},\ a_2=\dfrac{5s_A}{24},\ a_3=\dfrac{7s_A}{24},\ a_4=\dfrac{11s_A}{24}$ , i.e. $A=\{k,5k,7k,11k\}$ for a positive integer $k$ (with $s_A=24k$ ). For $(n,m)=(3,5)$ it gives $a_1=\dfrac{s_A}{60},\ a_2=\dfrac{11s_A}{60},\ a_3=\dfrac{19s_A}{60},\ a_4=\dfrac{29s_A}{60}$ , i.e. $A=\{k,11k,19k,29k\}$ (with $s_A=60k$ ). In both families one checks directly that $a_1+a_4=a_2+a_3=s_A/2$ and $a_1+a_3,a_1+a_2$ divide $s_A$ while $a_2+a_4,a_3+a_4$ lie strictly between $s_A/2$ and $s_A$ , so $n_A=4$ ; these are exactly the maximizing sets.