Find the value of minimize $a+b$

Question

If $a,b$ are positive integers and $$\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}$$. Find the value of minimize $a+b$

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From $\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}\Leftrightarrow13b^2-26b-12a=0$

$\Leftrightarrow 12(a+b)=13b^2-14b$ $\Leftrightarrow a+b=\frac{13b^2-14b}{12}$

$\Leftrightarrow a+b=b^2-b+\frac{b^2-2b}{12}=b^2-b+\frac{b\left(b-2\right)}{12}$

We have: $b$ must even number $(1)$

For $\frac{b\left(b-2\right)}{2.2.3}\in Z\Rightarrow b⋮3$ or $b-2⋮3$ $(2)$

From $(1);(2)$ $\Rightarrow b=6k$ or $b-2=6k$ $(k\ge1)$

*)With $b=6k\Rightarrow a+b=\frac{13\left(6k\right)^2-14\cdot 6k}{12}=3k^2-7k$

We have function $f(k)=39k^2-7k$ is function covariate withg $k\ge1$

So $a+b$ have minimize when k have minimize or $k=1$ $\Rightarrow b=6;a=26$

*)With $b-2=6k$.....

this method is inconvenient with me, so i can a new method

Answer 1

Hint:

$12a/13=b(b-2)$

$\implies13|a, a=13c$(say)

$(b-2)b=12c\implies$

$(i) b(b-2)\ge12\implies b>4$

$(ii)b$ must be even $b=2d, d>2$

$d(d-1)=3c$

Now $(d,d-1)=1$