Algebra Busters

By Cauchy-Schwarz inequality, $(k_{1}+k_{2}+...+k_{n})(\frac{1}{k_{1}}+\frac{1}{k_{2}}+...+\frac{1}{k_{n}}) \geq n^2$ so, we have $5n-4 \geq n^2$ . So, we have $(n-1)(n-4) \leq 0$ . This implies that $1 \leq n \leq 4$ .Equality holds for $n=1,4$ .From here, we will take 3 cases.

Case 1: $n=1,4$ .

It implies that $k_{1}=1$ is the only possibility for $n=1$ and $k_{1}=k_{2}=k_{3}=k_{4}=4$ is the only possibility for $n=4$ .

Case 2: $n=2$

It implies that $k_{1}+k_{2}=6, k_{1}k_{2}=6$ . So, $k_{1}, k_{2}$ are the roots of the equation $t^2-6t+6=0$ but this equation has no integer solutions.So no solution in this case.

Case 3: $n=3$

It implies that $k_{1}+k_{2}+k_{3}=11, \frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}=1$ . WLOG, we can assume that $k_{1} \leq k_{2} \leq k_{3}$ . Then doing case bashing, we will get that $(k_{1},k_{2},k_{3})$ are $(2,3,6)$ and its permutations.So, total number of solutions are $2+6=\boxed 8$

Consider the $n$ -dimensional vectors $\mathbf{a},\mathbf{b}$ where $a_j \; = \; \sqrt{k_j} \hspace{1cm} b_j \; = \; \frac{1}{\sqrt{k_j}} \hspace{2cm} 1 \le j \le n$ Then $||\mathbf{a}|| = \sqrt{5n-4}$ , $||\mathbf{b}|| = 1$ and $\mathbf{a}\cdot\mathbf{b} = n$ . By the Cauchy-Schwarz Inequality, $n \le \sqrt{5n-4}$ , which simplifies to read $(n-1)(n-4) \le 0$ , so that $1 \le n \le 4$ .

• If $n=1$ , we must have $k_1=1$ , leading the the singleton solution $(1)$ .
• If $n=2$ , the only solution of the equation $k_1^{-1} + k_2^{-1} = 1$ is $k_1=k_2=2$ , and $2 + 2 \neq 6$ . There are no solutions in this case.
• If $n=3$ , suppose for definiteness that $k_1 \le k_2 \le k_3$ . If $k_1 \ge 3$ then we must have $k_1=k_2=k_3=3$ , but since $3+3+3 \neq 11$ , this is not a solution. Thus we must have $k_1=2$ . The only solutions of $k_2^{-1} + k_3^{-1} = \tfrac12$ with $2 \le k_2 \le k_3$ are $k_2=k_3=4$ and $k_2=3,k_3=6$ . Only the second of these gives $k_1+k_2+k_3=11$ . Thus we have 6 more solutions: $(2,3,6),(2,6,3),(3,2,6),(3,6,2),(6,2,3),(6,3,2)$ .
• If $n=4$ , then the Cauchy=-Schwarz inequality at the start is an equality, so that $\mathbf{a}$ and $\mathbf{b}$ are proportional, and hence $k_1=k_2=k_3=k_4$ . Thus the only solution in this case is $(4,4,4,4)$ .

Thus there are $1 + 6 + 1 = \boxed{8}$ solutions.

Combining two rows in the following manner we get ,

$\sum_{i=1}^{n} K_i= 5n-4$ ........(a)

$\sum_{i=1}^{n} (\frac{5}{2})^2\frac{1}{K_i} = \frac{25}{4}$ ....(b)

Subtracting (b) from (a) and subtracting $5n$ from both sides of the new equation of (a)-(b) we get the following equation...

$\sum_{i=1}^{n} ({K_i} - 2*\frac{5}{2} + \frac{(2.5)^2}{K_i}) = \frac{25}{4} -4 = 2.25$

$\sum_{i=1}^{n}(\sqrt{K_i}-\frac{2.5}{\sqrt{K_i}})^2 = 2.25$ .........(1) And say, $A_i= (\sqrt{K_i}-\frac{2.5}{\sqrt{K_i}})^2$ .

$A_1=2.25 , A_2= 0.125 , A_3 =0.083333.... , A_4 =0.5625 , A_5 = 1.25, A_6 = 2.0416668$ and $A_K > 2.25$ for all $K\geq 7$ .

Now, we shall find the maximum how many times one natural number can be there in the solution tuple. Number of permitted appearance of some integer , $n(K)$ is as below

$\textbf {Upper bounds of K from (1)}$

{ $n(2)\leq{18} , n(3)\leq{27} ,n(4)\leq{4} ,n(5)\leq {1} , n(6)\leq{1} ,n(k)=0$ for all $k\geq{7}$ }

$\textbf {Upper bounf of k from "the reciprocal equation}"$

{ $n(2)\leq{2} , n(3)\leq{3} , n(4)\leq(4) ,n(5)\leq(5), n(6)\leq(6) ,....$ }

And from both of these, we get { $n(2)\leq{2} , n(3)\leq{3} , n(4)\leq(4) ,n(5)\leq(1), n(6)\leq(1), n(k\geq7)=0$ }...(2). But $5$ is of no use because it needs other multiples of $5$ to satisfy the reciprocal equation.

So, $K_i$ can only be $1,2,3,4,6$ . And from (2) we get $\sum_{1}^{n} K_i \leq 2*2+3*3+4*4+6={35} \Rightarrow n\leq7$ .

And from this little field we can easily find that the solutions exist for $n=1, n=3, n=4$ and these are {( $1$ )} ; { $(2,3,6)$ having $6$ permutations} ; { $(4,4,4,4)$ }. So, there are total 8 solutions.

Note : For $n=5,6,7$ we need terms like $4,6$ more than their available quantity and more $2,3$ to make the sum equal to $5n-4$ , but we can't use all the available $2,3$ because make the reciprocal sum more than $1$ , and overall there is a critical situation. So, $n$ can be only $1,3,4$ .