If (1/a) + (1/(2a)) + (1/(3a)) = (1/(b^2 − 2b)) a and b are positive integers Find minimum value of a + b


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Question Number 21236 by Joel577 last updated on 17/Sep/17

$If \frac{1}{a} + \frac{1}{2 a} + \frac{1}{3 a} = \frac{1}{b^{2} - 2 b}$ $a and b are positive integers$ $Find minimum value of a + b$
	Answered by mrW1 last updated on 17/Sep/17

	$\frac{1}{a} (1 + \frac{1}{2} + \frac{1}{3}) = \frac{1}{b (b - 2)}$ $\frac{11}{6 a} = \frac{1}{b (b - 2)}$ $a = \frac{11 b (b - 2)}{6} = 11 k$ $b (b - 2) = 6 k$ $b^{2} - 2 b - 6 k = 0$ $b = \frac{2 + \sqrt{4 + 4 \times 6 k}}{2} = 1 + \sqrt{1 + 6 k}$ $\Rightarrow k = 2 i (3 i \pm 1) with i = 1, 2, 3 . . . . . .$ $\Rightarrow b = 6 i + 1 \pm 1$ $\Rightarrow a = 22 i (3 i \pm 1)$ $k = 4, 8, 20, 28, 48, 60 . . .$ $b = 6, 8, 12, 14, 18, 20 . . .$ $a = 44, 88, 132, 154, 198, 220 . . .$ $k_{\min} = 4$ $b_{\min} = 1 + 5 = 6$ $a_{\min} = 11 \times 4 = 44$ $\Rightarrow {(a + b)}_{\min} = 50$
	Commented by $@ty@m last updated on 17/Sep/17

	$W i l l y o u p l . e x p l a i n b l u e p a r t ?$
	Commented by mrW1 last updated on 17/Sep/17

	$b = 1 + \sqrt{1 + 6 k}$ $1 + 6 k = j^{2}$ $k = \frac{j^{2} - 1}{6} = \frac{(j - 1) (j + 1)}{2 \times 3}$ $j must be odd, i . e . j = 2 m + 1$ $\Rightarrow k = \frac{2 m (2 m + 2)}{2 \times 3} = \frac{2 m (m + 1)}{3}$ $m must be 3 i or 3 i - 1$ $\Rightarrow k = 2 i (3 i + 1) or 2 i (3 i - 1)$ $\Rightarrow k = 2 i (3 i \pm 1)$ $b = 1 + \sqrt{1 + 6 k} = 1 + \sqrt{1 + 6 \times 2 i (3 i \pm 1)}$ $= 1 + \sqrt{{(6 i \pm 1)}^{2}}$ $= 1 + 6 i \pm 1$ $a = 11 k = 22 i (3 i \pm 1)$
	Commented by Joel577 last updated on 18/Sep/17

	$u n d e r s t o o d . t h a n k y o u v e r y m u c h$
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