Olympiad Problem 7

By Cauchy-Schwarz inequality, the product of the first two equations is at least $(1+1+...+1)^{2}=n^{2}$

Hence $n^{2}-5n+4$ is less than 0. Solving the quadratic inequality, $n$ is between 1 and 4.

If $n=1$ , then $k_{1}=1$

If $n=2$ , then $k_{1}+k_{2}=6$ . Trying all pairs (there are only 3), none work.

If $n=3$ , then the sum of reciprocals of $k_{1}, k_{2}, k_{3}$ is 1. Suppose, without loss of generality, that $k_{1}$ is the least among these three numbers. Then suppose it is more than 3. The sum would be less than $\frac {1}{3} \times 3=1$ . So it is 2 or 3. Checkig each case, (2, 3, 6), (2, 4, 4) and (3, 3, 3) are the only solutions (ignoring order) and only (2, 3, 6) is possible.

If $k=4$ , it $5n-4=n^{2}$ and it becomes the equality case of Cauchy-Schwarz inequality, hence all $k_{i}$ , are equal. Hence (4, 4, 4, 4) is the only solution. Adding up all, the answer is 36.

(Actually the question is quite unclear as we don't know if order counts)

Titu's lemma on second equation gives

$\sum_{i=1}^n k_i \geq n^2$

so combining with the first equation we get

$n^2 -5n + 4 \leq 0$

This means $n \in \{1, 2, 3, 4\}$

For $n = 1$ we immediately get $k_1 = 1$ works as the unique solution so we have the integers $n = 1, k_1 = 1$ as first solutions.

For $n = 2$ we search the solutions betweek the couple $k_1, k_2$ that sums up to $6$ .

Notice that $1$ cannot belong to any $k$ otherwise we have the sum of the reciprocal is strictly greater than $1$ .

Also we can optimize the checks noting that in a couple of solutions we have $k_{min} \leq 3$ and $k_{max} \geq 3$ .

So we have to check

$k_1 = 2, k_2 = 4$ that fails

$k_1 = 3, k_2 = 3$ that fails.

No set of solutions for the case $n = 2$ .

For $n = 3$ we search the solutions betweek the couple $k_1, k_2, k_3$ that sums up to $11$ .

Notice that $1$ cannot belong to any $k$ otherwise we have the sum of the reciprocal is strictly greater than $1$ .

Also no two values of $k$ can be equal to $2$ for the same reason.

Also we can optimize the checks noting that in a couple of solutions we have $k_{min} \leq 3$ and $k_{max} \geq 4$ .

So we have to check

$k_1 = 2, k_2 = 3, k_3 = 6$ that works

$k_1 = 2, k_2 = 4, k_3 = 5$ that fails

$k_1 = 3, k_2 = 3, k_3 = 5$ that fails

$k_1 = 3, k_2 = 4, k_3 = 4$ that fails.

So we discover the integers $n = 3, k_1 =2, k_2 =3, k_3=6$ as second set of solution

For $n = 4$ we search the solutions betweek the couple $k_1, k_2, k_3, k_4$ that sums up to $16$ .

Notice that $1$ cannot belong to any $k$ otherwise we have the sum of the reciprocal is strictly greater than $1$ .

Also no two values of $k$ can be equal to $2$ for the same reason.

Also no three values of $k$ can be equal to $3$ for the same reason.

Also we can optimize the checks noting that in a couple of solutions we have $k_{min} \leq 4$ and $k_{max} \geq 4$ .

So we have to check

$k_1 = 2, k_2 =3 , k_3 =3 , k_4= 8$ that fails

$k_1 = 2, k_2 = 3, k_3 = 4, k_4= 7$ that fails

$k_1 = 2, k_2 = 3, k_3 = 5, k_4= 6$ that fails

$k_1 = 2, k_2 = 4, k_3 = 4, k_4= 6$ that fails

$k_1 = 2, k_2 = ,4 k_3 = 5, k_4= 5$ that fails

$k_1 = 3, k_2 = 3, k_3 = 4, k_4= 6$ that fails

$k_1 = 3, k_2 = 3, k_3 = 5, k_4= 5$ that fails

$k_1 = 3, k_2 = 4, k_3 = 4, k_4= 5$ that fails

$k_1 = 4, k_2 = 4, k_3 = 4, k_4= 4$ that works.

So we discover the integers $n = 4, k_1 =4, k_2 =4, k_3=4, k_4=4$ as a third and last set of solution.

Summing the sets of solutions we get the answer $36$ .