Hardest Multiple Ever!

Relevant wiki: Diophantine Equations - Solve by Factoring

Mr Mark Hennings caught my mistake and I've amended it.

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Let's start with the last equation: $\dfrac {abc}{ b- \dfrac1{c-\dfrac1a}} = 4 \Leftrightarrow abc = 4\left( \dfrac{abc-a-b}{ac-1} \right) \Leftrightarrow abc(ac-1)=4(abc-a-b)\qquad \qquad (1)$

Since $(bc-1)(ac-1) - c(abc - a - b) \; = \; 1$ , you have that $ac-1$ and $abc - a - b$ are coprime; since $ac-1$ divides $4(abc - a - b)$ we can now deduce that $ac-1$ divides $4$ .

So $ac - 1$ must be equal to $-1,1, -2,2, -4$ or $4$ , or equivalently $ac = 0,2,-1,3,-3$ or $5$ .

But if $ac = 0$ , then $a$ and/or $c$ is equal to $0$ , which is impossible, because the LHS of any of the three given equations is equal to $0$ , thus $ac \ne 0$ .

We are left with 5 cases to check: $ac = -3,-1,2,3,5$ .

Case 1 : Suppose $ac=-3$ , substituting it into $(1)$ gives $(-3b)(-3-1) = 4(-3b - a-b) \Leftrightarrow 7b = -a \; .$ This tells us that $a$ is divisible by 7, but since $ac = -3$ , then $a$ can only take values of $\pm1,\pm3$ , which contradicts our findings. So, $ac = -3$ is not a solution.

Case 2 : Suppose $ac = -1$ , substituting it into $(1)$ gives $(-1b)(-1-1) = 4(-b-a-b) \Leftrightarrow 5b = -2a\; .$ This tells us that $a$ is divisible by 5, but since $ac = -1$ , then $a$ can only take values of $\pm1,$ , which contradicts our findings. So, $ac = -1$ is not a solution.

Case 3 : Suppose $ac = 2$ , substituting it into $(1)$ gives $(2b)(2-1) = 4(2b-a-b) \Leftrightarrow b=2a\; .$

Since $ac = 2$ and $b = 2a$ , then one of the possible solutions of $(a,b,c)$ are $(1,2,2), (-1,-2,-2), (2,4,1), (-2,-4,-1)$ . By trial and error, we can see that none of any of these solutions both the other two given equations.

Case 4 : Suppose $ac = 3$ , substituting it into $(1)$ gives $(3b)(3-1) = 4(3b-a-b) \Leftrightarrow b=2a\; .$

Since $ac = 3$ and $b = 2a$ , then one of the possible solutions of $(a,b,c)$ are $(1,2,3), (-1,-2,-3), (3,4,1), (-3,-4,-1)$ . By trial and error, we can see that only two of these solution satisfy the other two given equations, namely $(a,b,c) = (-1,-2,-3), (1,2,3)$ .

Case 5 : Suppose $ac = 5$ , substituting it into $(1)$ gives $(5b)(5-1) = 4(5b-a-b) \Leftrightarrow a=-b\; .$

Since $ac = 5$ and $a = -b$ , then one of the possible solutions of $(a,b,c)$ are $(1,-1,5), (-1,1,-5),(-5,5,-1),(5,-5,1)$ . By trial and error, we can see that none of any of these solutions both the other two given equations.

From all these cases, only Case 4 yields a possible solution, namely $(a,b,c) = (1,2,3), (-1,-2,-3)$ . Either way, $\text{lcm}(a,b,c) = \boxed6$ .