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If p , q p, q and r r are all prime numbers greater than 3, what is the largest number which always divides p 2 q 2 r 2 p 2 q 2 q 2 r 2 p 2 r 2 + p 2 + q 2 + r 2 1 p^2q^2r^2 - p^2q^2 - q^2r^2 - p^2r^2 + p^2 + q^2 + r^2 - 1 ?


The answer is 13824.

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1 solution

Muhammad Ihsan
May 9, 2015

f a c t o r i z e ( p 2 1 ) ( q 2 1 ) ( r 2 1 ) w e k n o w t h a t a l l p r i m e n u m b e r c a n b e w r i t e a s 6 k ± 1 . l e t , p = 6 k ± 1 q = 6 m ± 1 r = 6 n ± 1 . s o , t h a t i t c a n b e w r i t t e n 12 3 k m n ( 3 k ± 1 ) ( 3 m ± 1 ) ( 3 n ± 1 ) . k = m = n = 1 i s t h e s m a l l e s t v a l u e . s o , t h e g r e a t e s t d i v i d e r i s 12 3 2 3 = 13824 factorize\quad \\ ({ p }^{ 2 }-1)({ q }^{ 2 }-1)({ r }^{ 2 }-1)\\ we\quad know\quad that\quad all\quad prime\quad number\quad can\quad be\quad write\quad as\quad 6k\pm 1\\ .\\ let,\\ p=6k\pm 1\\ q=6m\pm 1\\ r=6n\pm 1\\ .\\ so,\quad that\quad it\quad can\quad be\quad written\\ { 12 }^{ 3 }kmn(3k\pm 1)(3m\pm 1)(3n\pm 1)\\ .\\ k=m=n=1\quad is\quad the\quad smallest\quad value.\\ so,\quad the\quad greatest\quad divider\quad is\quad { 12 }^{ 3 }{ 2 }^{ 3 }=13824\\ \\

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