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For all positive integers $n$, $14^{6n} - 11^{6n}$ is divisible by ?

This question is followed with four options:

$1)157\quad\quad 2) 163\quad\quad 3) 225\quad\quad \quad 4) \text{All of these}$

I don't know how to do this in a "fast" way manually,what I did is used computer and factorized $14^6-11^6$ which gives $5757975=157\times 163 \times 225$ but surely this is not what I am supposed to do during exams as it will be too tedious even if I check for divisibility of the three numbers manually using the options but does computing $14^6-11^6$ and then checking for divisibility is the best option for solving this problem (manually/using pencil and paper)?

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You could at least speed it up a little by noting that $14^6-11^6 = (14^3-11^3)\cdot (14^3+11^3) = 1413 \cdot 4075$, so that you only have to factor those two smaller numbers to see which primes divide $14^{6n}-11^{6n}$. –  TMM Sep 19 '11 at 19:42
    
@Thijs Laarhoven:Aha,that will help a bit :) –  Quixotic Sep 19 '11 at 19:43
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Furthermore, $(14^3 \pm 11^3) = (14 \pm 11)(14^2 \mp 14 \cdot 11 + 11^2)$ which reduces the complexity of the problem further. –  JavaMan Sep 19 '11 at 19:44
    
@DJC:Indeed! :) –  Quixotic Sep 19 '11 at 19:47
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@jspecter: Factorization is much faster than computing $15^6-11^6$ by hand: you get $3\cdot 25(14^2+14\cdot 11+11^2)(14^2-14\cdot 11+11^2)=3\cdot 25(196+275)(196-33)=$ $3\cdot 25\cdot 471\cdot 163=3^2\cdot 5^2\cdot 157\cdot 163=225\cdot 157\cdot 163$ by a calculation whose individual steps qualify as mental arithmetic. –  Brian M. Scott Sep 19 '11 at 20:13
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1 Answer

up vote 8 down vote accepted

HINT $\rm\ x^6 - y^6\ =\ (x-y)\ (x+y)\ (x^2+y^2-xy)\ (x^2+y^2+xy)$

$\qquad\qquad\qquad\qquad\qquad =\ 3\cdot 5^2\ (317-154)\ (317+154)\ =\ 3\cdot 5^2\cdot 163\ ( 3\cdot 157)\ \ $ for $\rm\:x,y = 14,11$

Since $\ 3^2\cdot 5^2\: =\: 15^2\: =\: 225\:,\ $ we deduce $\rm\:157,\:163,\: 225\ |\ 14^6-11^6\ |\ 14^{\:6\:n}-11^{\:6\:n}\:.$

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