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What is the smallest prime factor of 1 9 n 8 n 19^{n}-8^{n} , where 2 n 2|n ?


The answer is 3.

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1 solution

Finn Hulse
May 1, 2014

Given that n n is even, let's define k = n 2 k=\frac{n}{2} . We can use difference of squares to get ( 1 9 k 8 k ) ( 1 9 k + 8 k ) (19^{k}-8^{k})(19^k+8^k) . We can assume WLOG that k = 1 k=1 . Thus, we achieve ( 11 ) ( 27 ) (11)(27) , the smallest prime factor of which is simply 3 \boxed{3} .

How is the assumption WLOG?

Kenny Lau - 6 years, 11 months ago

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