During my idle days

Number Theory Level 3

Let $x, y, z$ be three positive integers with $x<y<z$ .

$\frac 1{xy}+\frac 1{yz}+\frac 1{zx}=\frac 14$

If there are $n$ solutions to the equation above of the form $(x_1,y_1,z_1)$ , $(x_2,y_2,z_2)$ , $(x_3,y_3,z_3)$ , $\cdots (x_n,y_n,z_n)$ , what is the value of $\displaystyle \sum_{k=1}^n (x_k+y_k+z_k)$ ?

The answer is 96.

1 solution

Qweros Bistoros
Apr 8, 2020

Rewrite eqauation as $x+y+z=\frac{xyz}{4}$

If $x≥3$ then $y≥4$ , so $\frac{xyz}{4}≥3z$ , but $x+y+z<3z$

If $x=1$ then $1+y+z=\frac{yz}{4}$ , so $z=4+\frac{20}{y-4}$ which is integer and grater than y for $y \in \{5, 6, 8\}$

If $x=2$ , then $2+y+z=\frac{yz}{2}$ , so $z=2+\frac{8}{y-2}$ which is integer and grater than y for $y \in \{ 3, 4 \}$

So solutions are $(1,5,24), (1,6,14), (1,8,9), (2,3,10), (2,4,6)$

During my idle days

The answer is 96.

1 solution

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