Determine number/s that is/are comprised of four distinct prime factors such that difference of largest and smallest prime factors is equal to the sum of remaining two factors.


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Question Number 9025 by Rasheed Soomro last updated on 15/Nov/16

$Determine number / s that is / are comprised$ $of four distinct prime factors such that$ $difference of largest and smallest prime$ $factors is equal to the sum of remaining$ $two factors ._{Propsed by Rasheed Soomro}$
Commented by FilupSmith last updated on 15/Nov/16
$Do you mean: n=p_1 p_2 p_3 p_4 p_a >p_(a+1) p_1 −p_4 =p_2 +p_3 p_a ∈P, p_a ≠p_(a+i) ∀i/{0} n=?$
$Do you mean :$ $n = p_{1} p_{2} p_{3} p_{4}$ $p_{a} > p_{a + 1}$ $p_{1} - p_{4} = p_{2} + p_{3}$ $p_{a} \in P, p_{a} \neq p_{a + i} \forall i / {0}$ $n = ?$
Commented by Rasheed Soomro last updated on 16/Nov/16

$Yes Filup I mean that .$
Commented by FilupSmith last updated on 16/Nov/16

$w o r k i n g / p l a y i n g a r o u n d$ $x = p_{1} p_{2} p_{3} p_{4}$ $p_{1} - p_{4} = p_{2} + p_{3}$ $p_{1} = p_{2} + p_{3} + p_{4}$ $∴ x = (p_{2} + p_{3} + p_{4}) p_{2} p_{3} p_{4}$ $∴ p_{2} + p_{3} + p_{4} \in P$ $p_{2} + p_{3} + p_{4} < p_{2} < p_{3} < p_{4}$ $p_{2} < p_{2} - p_{3} < - p_{4} < - p_{3}$ $p_{2} is + ve$ $p_{2} - p_{3} is - ve$ $- p_{4} and - p_{3} are - ve$ $∴ i d o n t t h i n k s u c h n u m b e r e x i s t s$ $a s t h e e q u a l i t y i s n o t p o s s i b l e$ $w o r k i n g / p l a y i n g a r o u n d$
Commented by mrW last updated on 16/Nov/16

$w h e n I t a k e a l o o k a t t h e t a b l e o f$ $p r i m e n u m b e r s I t h i n k t h e r e$ $s h o u l d b e i n f i n i t e l y m a n y s u c h$ $n u m b e r s, s u c h a s$ $x = 3 \times 5 \times 11 \times 19$ $x = 5 \times 7 \times 11 \times 23$ $x = 3 \times 7 \times 13 \times 23$ $e t c .$
Commented by Rasheed Soomro last updated on 16/Nov/16

$^{∙} TH α n k s o f y o u b o t h!$ $^{∙} Good attack of Filup to the problem but$ $failed in the end!$ $^{∙} I also think that such numbers are infinite$ $but Can we prove that such numbers are infimite ?$
Commented by FilupSmith last updated on 16/Nov/16

$I wonder where I went wrong$
Commented by FilupSmith last updated on 16/Nov/16

$I think my mistake was that in my m ath,$ $p_{1} is actually > p_{2}, but i missthought$ $it as p_{1} < p_{2}, or vica versa .$ $Everything up to$ $x = (p_{2} + p_{3} + p_{4}) p_{2} p_{3} p_{4}$ $is correct . BUT,$ $it should be that :$ $p_{1} > p_{2} > p_{3} > p_{3} .$
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