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prove that isn't prime

Prove that \(14^{n}+ 11\) is never prime

Note by A K
2 years, 11 months ago

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I am assuming that you wish this proved for integers \(n \ge 1\), (for \(14^{0} + 11 = 12\) is not prime).

If \(n\) is odd then the last digit of \(14^{n}\) will be \(4\) and thus the last digit of \(14^{n} + 11\) will be \(5\), implying that \(14^{n} + 11\) is divisible by \(5\).

If \(n\) is even then let \(n = 2k\) for some integer \(k\). Then \(14^{n} + 11 = 196^{k} + 11 = (195 + 1)^{k} + 11\), (A).

Now in the binomial expansion of \((195 + 1)^{k}\) we will have a factor of \(195\) in every term except for the term \(1^{k} = 1\). We can then write equation (A) as \(1 + 195*m + 11 = 12 + 195*m = 3*(4 + 65*m)\) for some integer \(m\). Thus in this case \(14^{n} + 11\) is divisible by \(3\).

So for all integers \(n \ge 1\) we can conclude that \(14^{n} + 11\) is not prime.

Brian Charlesworth - 2 years, 11 months ago

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Great proof! Although I think for the last half, you can look at the equation mod 3, and this get \((-1)^n-1 \pmod{3}\) which for all even n, we get \(1-1=0 \pmod{3}\). It's similar to your method but just a tad simplier.

Trevor Arashiro - 2 years, 11 months ago

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Good point. Also, for the first part we could have looked at the expression mod 5, to get \((-1)^{n} + 1 \equiv 0 \mod{5}\) for all odd n.

It takes a bit of intuition to guess at which mod values to 'filter' with, but it makes for a more elegant approach. :)

Brian Charlesworth - 2 years, 11 months ago

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14 gives remainder 2 when divided by 3.So,when n=1,2,3,4... , \(14^n\) will give remainder 2,1,2,1 .... So every even power gives remainder 1. This means 14^\(n\)+11 will be of the form 3m+12 which cant be a prime. 14 gives remainder 4 when divided by 5.when n=1,2,3,4...\(14^n\) will give rem. 4,1,4,1.. So every odd power gives remainder 4. this means \(14^n\)+11 will be of the form 5m+15 which cant be prime.

Rutwik Dhongde - 2 years, 10 months ago

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