# prove that isn't prime

Prove that $$14^{n}+ 11$$ is never prime

Note by A K
3 years, 7 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.

- 3 years, 7 months ago

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.

- 3 years, 7 months ago

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. :)

- 3 years, 7 months ago

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.

- 3 years, 6 months ago